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街头斗殴数学

Street-Fighting Mathematics

街头斗殴数学

Street-Fighting Mathematics



有根据的猜测和机会主义的问题解决的艺术

The Art of Educated Guessing and

Opportunistic Problem Solving

桑乔伊·马哈詹

Sanjoy Mahajan

卡弗·A·米德作序

Foreword by Carver A. Mead

麻省理工学院出版社

马萨诸塞州剑桥

英国伦敦

The MIT Press

Cambridge, Massachusetts

London, England

© 2010 Sanjoy Mahajan

© 2010 by Sanjoy Mahajan

前言 © 2010 Carver A. Mead

Foreword © 2010 by Carver A. Mead

图片

《街头斗殴数学:有根据的猜测和机会主义问题解决的艺术》由 Sanjoy Mahajan(作者)、Carver A. Mead(序言)和麻省理工学院出版社(出版商)编写,根据知识共享署名-非商业性-相同方式共享 3.0 美国许可证授权。

Street-Fighting Mathematics: The Art of Educated Guessing and Opportunistic Problem Solving by Sanjoy Mahajan (author), Carver A. Mead (foreword), and MIT Press (publisher) is licensed under the Creative Commons Attribution–Noncommercial–Share Alike 3.0 United States License.

许可证副本可在http://creativecommons.org/licenses/by-nc-sa/3.0/us/获取。

A copy of the license is available at http://creativecommons.org/licenses/by-nc-sa/3.0/us/

有关特殊数量折扣的信息,请发送电子邮件至 special_sales@mitpress.mit.edu

For information about special quantity discounts, please email special_sales@mitpress.mit.edu

作者使用 ConT E Xt 和 PDFT E X在 Palatino 和 Euler 中排版

Typeset in Palatino and Euler by the author using ConTEXt and PDFTEX

国会图书馆在版数据

LIBRARY OF CONGRESS CATALOGING-IN-PUBLICATION DATA

马哈詹,桑乔伊,1969–

Mahajan, Sanjoy, 1969–

街头斗殴数学:有根据的猜测和机会主义问题解决的艺术 / Sanjoy Mahajan;Carver A. Mead 作序。

Street-fighting mathematics : the art of educated guessing and opportunistic problem solving / Sanjoy Mahajan ; foreword by Carver A. Mead.

     页厘米

     p. cm.

包括参考书目和索引。

Includes bibliographical references and index.

ISBN 978-0-262-51429-3(平装:碱性纸)

ISBN 978-0-262-51429-3 (pbk. : alk. paper)

ISBN 978-0-262-26559-1(电子书)

ISBN 978-0-262-26559-1 (e-book)

1. 问题解决。2. 假设。3. 估计理论。I. 标题。

1. Problem solving. 2. Hypothesis. 3. Estimation theory. I. Title.

QA63.M34 2010

QA63.M34 2010

510—dc22

510—dc22

2009028867

2009028867

美国印刷装订

Printed and bound in the United States of America

10 9 8 7 6 5 4 3 2 1

10  9  8  7  6  5  4  3  2  1

致朱丽叶

For Juliet

简要内容

Brief contents

前言

Foreword

前言

Preface

1 尺寸

1    Dimensions

2个简单案例

2    Easy cases

3 合并

3    Lumping

4 图片证明

4    Pictorial proofs

5 取出大部件

5    Taking out the big part

6 类比

6    Analogy

参考书目

Bibliography

指数

Index

内容

Contents

前言

Foreword

前言

Preface

1 尺寸

1    Dimensions

1.1 经济学:跨国公司的力量

1.1 Economics: The power of multinational corporations

1.2 牛顿力学:自由落体

1.2 Newtonian mechanics: Free fall

1.3 猜测积分

1.3 Guessing integrals

1.4总结和进一步的问题

1.4 Summary and further problems

2个简单案例

2    Easy cases

2.1 重温高斯积分

2.1 Gaussian integral revisited

2.2 平面几何:椭圆的面积

2.2 Plane geometry: The area of an ellipse

2.3 立体几何:截头金字塔的体积

2.3 Solid geometry: The volume of a truncated pyramid

2.4 流体力学:阻力

2.4 Fluid mechanics: Drag

2.5总结和进一步的问题

2.5 Summary and further problems

3 合并

3    Lumping

3.1 估计人口:有多少婴儿?

3.1 Estimating populations: How many babies?

3.2 估算积分

3.2 Estimating integrals

3.3 估计导数

3.3 Estimating derivatives

3.4 分析微分方程:弹簧质量系统

3.4 Analyzing differential equations: The spring–mass system

3.5 预测单摆的周期

3.5 Predicting the period of a pendulum

3.6总结和进一步的问题

3.6 Summary and further problems

4 图片证明

4    Pictorial proofs

4.1 奇数加法

4.1 Adding odd numbers

4.2 算术平均数和几何平均数

4.2 Arithmetic and geometric means

4.3 近似对数

4.3 Approximating the logarithm

4.4 平分三角形

4.4 Bisecting a triangle

4.5 求和级数

4.5 Summing series

4.6总结和进一步的问题

4.6 Summary and further problems

5 取出大部件

5    Taking out the big part

5.1 使用“一”和“少”进行乘法

5.1 Multiplication using one and few

5.2 分数变化和低熵表达式

5.2 Fractional changes and low-entropy expressions

5.3 具有一般指数的分数变化

5.3 Fractional changes with general exponents

5.4 逐次逼近:井有多深?

5.4 Successive approximation: How deep is the well?

5.5 令人畏惧的三角积分

5.5 Daunting trigonometric integral

5.6总结和进一步的问题

5.6 Summary and further problems

6 类比

6    Analogy

6.1 空间三角学:甲烷中的键角

6.1 Spatial trigonometry: The bond angle in methane

6.2 拓扑:有多少个区域?

6.2 Topology: How many regions?

6.3 算子:欧拉-麦克劳林求和

6.3 Operators: Euler–MacLaurin summation

6.4 正切根:一个令人畏惧的超越和

6.4 Tangent roots: A daunting transcendental sum

6.5一路顺风

6.5 Bon voyage

参考书目

Bibliography

指数

Index

前言

Foreword

我们大多数人都从数学家那里学习数学课程——坏主意!

Most of us took mathematics courses from mathematicians—Bad Idea!

数学家视数学为一门独立的学科。我们其他人则将数学作为一种精确的语言,用来表达现实世界中量之间的关系,并以此作为从这些关系中得出定量结论的工具。然而,如今教授的数学课程却很少能起到这样的作用,甚至常常带来负面影响。

Mathematicians see mathematics as an area of study in its own right. The rest of us use mathematics as a precise language for expressing relationships among quantities in the real world, and as a tool for deriving quantitative conclusions from these relationships. For that purpose, mathematics courses, as they are taught today, are seldom helpful and are often downright destructive.

作为一名学生,我曾承诺过自己,如果我将来成为一名老师,我绝不会让学生接受那种教学方式。我一生都在努力寻找直接透明地观察现实的方法,并试图将这些洞见量化地表达出来,我从未故意违背过我的诺言。

As a student, I promised myself that if I ever became a teacher, I would never put a student through that kind of teaching. I have spent my life trying to find direct and transparent ways of seeing reality and trying to express these insights quantitatively, and I have never knowingly broken my promise.

除了极少数例外,我发现最有用的数学知识都是在科学和工程课上、自学或从本书中学到的。《街头斗殴数学》是一股清风。Sanjoy Mahajan 以最友好的方式向我们传授了在现实世界中行之有效的工具。就在我们认为某个主题显而易见的时候,他又将我们带入了另一个层次。我个人最喜欢的是纳维-斯托克斯方程的解法:它太难了,我甚至从未尝试过求解。但他带领我们解决了这个问题,并一路收获了宝贵的洞见。

With rare exceptions, the mathematics that I have found most useful was learned in science and engineering classes, on my own, or from this book. Street-Fighting Mathematics is a breath of fresh air. Sanjoy Mahajan teaches us, in the most friendly way, tools that work in the real world. Just when we think that a topic is obvious, he brings us up to another level. My personal favorite is the approach to the Navier–Stokes equations: so nasty that I would never even attempt a solution. But he leads us through one, gleaning gems of insight along the way.

这本小书里蕴含着我们每个人的真知灼见。本书中的一些技巧,我亲自实践过,也推荐给大家。

In this little book are insights for every one of us. I have personally adopted several of the techniques that you will find here. I recommend it highly to every one of you.

—卡弗·米德

—Carver Mead

前言

Preface

过度的数学严谨性会导致僵硬:即使得出了正确的结果,也害怕做出不合理的猜测。与其麻痹大意,不如鼓起勇气——先行动,后提问。虽然作为公共政策并不明智,但这却是一种宝贵的解决问题的哲学,也是本书的主题:如何在没有证明或精确计算的情况下猜测答案。

Too much mathematical rigor teaches rigor mortis: the fear of making an unjustified leap even when it lands on a correct result. Instead of paralysis, have courage—shoot first and ask questions later. Although unwise as public policy, it is a valuable problem-solving philosophy, and it is the theme of this book: how to guess answers without a proof or an exact calculation.

有根据的猜测和随机应变的问题解决需要一个工具箱。用乔治·波利亚的话来说,工具是我使用过两次的技巧。本书构建、完善并演示了跨人类知识不同领域的实用工具。丰富的示例有助于将工具(一般原理)与具体应用区分开来,以便你能够掌握工具并将其迁移到你感兴趣的问题中。

Educated guessing and opportunistic problem solving require a toolbox. A tool, to paraphrase George Polya, is a trick I use twice. This book builds, sharpens, and demonstrates tools useful across diverse fields of human knowledge. The diverse examples help separate the tool—the general principle—from the particular applications so that you can grasp and transfer the tool to problems of particular interest to you.

用于教授该工具的例子包括:不积分地猜测积分、反驳媒体中的常见论点、从非线性微分方程中提取物理性质、不解纳维-斯托克斯方程地估计阻力、寻找平分三角形的最短路径、猜测键角、以及对每个项都是未知和超越的无穷级数求和。

The examples used to teach the tools include guessing integrals without integrating, refuting a common argument in the media, extracting physical properties from nonlinear differential equations, estimating drag forces without solving the Navier–Stokes equations, finding the shortest path that bisects a triangle, guessing bond angles, and summing infinite series whose every term is unknown and transcendental.

本书是对《如何解决问题》 [ 37 ]、《数学与合理推理》 [ 35,36 ]和问题解决的艺术与技巧》 [ 49 ]等著作的补充。这些著作教导我们如何精确地解决精确陈述的问题,而生活中我们经常遇到一些定义不明确的问题,只需要中等精度的解决方案。精度仅为2倍的计算可能表明,拟建的桥梁永远不会建成,或者电路永远无法正常工作。省去进行精确分析所节省的精力,可以用来发明有前景的新设计。

This book complements works such as How to Solve It [37], Mathematics and Plausible Reasoning [35, 36], and The Art and Craft of Problem Solving [49]. They teach how to solve exactly stated problems exactly, whereas life often hands us partly defined problems needing only moderately accurate solutions. A calculation accurate only to a factor of 2 may show that a proposed bridge would never be built or a circuit could never work. The effort saved by not doing the precise analysis can be spent inventing promising new designs.

本书源于我在麻省理工学院教授的同名短期课程,该课程持续了数年。学生的经历各不相同:从本科一年级到准备从事研究和教学工作的研究生。学生的专业领域也各不相同:从物理、数学、管理学到电气工程、计算机科学和生物学,应有尽有。尽管如此,或者正因为如此,学生们似乎受益于这套工具,并享受着插图和应用的多样性。祝愿你们也一样。

This book grew out of a short course of the same name that I taught for several years at MIT. The students varied widely in experience: from first-year undergraduates to graduate students ready for careers in research and teaching. The students also varied widely in specialization: from physics, mathematics, and management to electrical engineering, computer science, and biology. Despite or because of the diversity, the students seemed to benefit from the set of tools and to enjoy the diversity of illustrations and applications. I wish the same for you.

如何使用本书

How to use this book

亚里士多德曾是年轻的马其顿国王亚历山大(后来的亚历山大大帝)的导师。正如古代皇室所知,一位技艺精湛、知识渊博的导师是最有效的教师[ 8 ]。一位技艺精湛的导师少言寡语,多提问,因为她深知提问、思考和讨论能够促进持久的学习。因此,本书穿插了两种类型的问题。

Aristotle was tutor to the young Alexander of Macedon (later, Alexander the Great). As ancient royalty knew, a skilled and knowledgeable tutor is the most effective teacher [8]. A skilled tutor makes few statements and asks many questions, for she knows that questioning, wondering, and discussing promote long-lasting learning. Therefore, questions of two types are interspersed through the book.

图片页边空白处标有 的问题:这些问题是导师在辅导课上可能会问你的,并要求你制定分析的后续步骤。这些问题的答案在后续文本中,你可以在其中查看你的答案和我的分析。

Questions marked with a in the margin: These questions are what a tutor might ask you during a tutorial, and ask you to work out the next steps in an analysis. They are answered in the subsequent text, where you can check your solutions and my analysis.

编号问题:这些问题用阴影背景标记,是导师在辅导课后可能会给你带回家的。它们要求你练习工具,扩展示例,同时使用多个工具,甚至解决(明显的)悖论。

Numbered problems: These problems, marked with a shaded background, are what a tutor might give you to take home after a tutorial. They ask you to practice the tool, to extend an example, to use several tools together, and even to resolve (apparent) paradoxes.

尝试两种类型的许多问题!

Try many questions of both types!

版权许可

Copyright license

本书采用与麻省理工学院开放课程相同的许可协议:知识共享署名-非商业性使用-相同方式共享许可协议。出版商和我鼓励您以非商业形式使用、改进和分享本书,我们乐意接受任何更正和建议。

This book is licensed under the same license as MIT’s OpenCourseWare: a Creative Commons Attribution-Noncommercial-Share Alike license. The publisher and I encourage you to use, improve, and share the work non-commercially, and we will gladly receive any corrections and suggestions.

致谢

Acknowledgments

我衷心感谢以下个人和组织。

I gratefully thank the following individuals and organizations.

标题:卡尔·莫耶 (Carl Moyer)。

For the title: Carl Moyer.

编辑指导: Katherine Almeida 和 Robert Prior。

For editorial guidance: Katherine Almeida and Robert Prior.

对手稿进行全面、透彻的评论: Michael Gottlieb、David Hogg、David MacKay 和 Carver Mead。

For sweeping, thorough reviews of the manuscript: Michael Gottlieb, David Hogg, David MacKay, and Carver Mead.

为激励大家而设的老师有: John Allman、Arthur Eisenkraft、Peter Goldreich、John Hopfield、Jon Kettenring、Geoffrey Lloyd、Donald Knuth、Carver Mead、David Middlebrook、Sterl Phinney 和 Edwin Taylor。

For being inspiring teachers: John Allman, Arthur Eisenkraft, Peter Goldreich, John Hopfield, Jon Kettenring, Geoffrey Lloyd, Donald Knuth, Carver Mead, David Middlebrook, Sterl Phinney, and Edwin Taylor.

感谢以下人员提出的许多宝贵建议和讨论: Shehu Abdussalam、Daniel Corbett、Dennis Freeman、Michael Godfrey、Hans Hagen、Jozef Hanc、Taco Hoekwater、Stephen Hou、Kayla Jacobs、Aditya Mahajan、Haynes Miller、Elisabeth Moyer、Hubert Pham、Benjamin Rapoport、Rahul Sarpeshkar、Madeleine Sheldon-Dante、Edwin Taylor、Tadashi Tokieda、Mark Warner 和 Joshua Zucker。

For many valuable suggestions and discussions: Shehu Abdussalam, Daniel Corbett, Dennis Freeman, Michael Godfrey, Hans Hagen, Jozef Hanc, Taco Hoekwater, Stephen Hou, Kayla Jacobs, Aditya Mahajan, Haynes Miller, Elisabeth Moyer, Hubert Pham, Benjamin Rapoport, Rahul Sarpeshkar, Madeleine Sheldon-Dante, Edwin Taylor, Tadashi Tokieda, Mark Warner, and Joshua Zucker.

有关写作过程的建议: Carver Mead 和 Hillary Rettig。

For advice on the process of writing: Carver Mead and Hillary Rettig.

有关书籍设计的建议: Yasuyo Iguchi。

For advice on the book design: Yasuyo Iguchi.

有关免费许可的建议: Daniel Ravicher 和 Richard Stallman。

For advice on free licensing: Daniel Ravicher and Richard Stallman.

对于用于计算的免费软件: Fredrik Johansson(mpmath)、Maxima 项目和 Python 社区。

For the free software used for calculations: Fredrik Johansson (mpmath), the Maxima project, and the Python community.

对于用于排版的自由软件: Hans Hagen 和 Taco Hoekwater (ConT E Xt);Han The Thanh (PDFT E X);Donald Knuth (T E X);John Hobby (MetaPost);John Bowman、Andy Hammerlindl 和 Tom Prince (Asymptote);Matt Mackall (Mercurial);Richard Stallman (Emacs);以及 Debian GNU/Linux 项目。

For the free software used for typesetting: Hans Hagen and Taco Hoekwater (ConTEXt); Han The Thanh (PDFTEX); Donald Knuth (TEX); John Hobby (MetaPost); John Bowman, Andy Hammerlindl, and Tom Prince (Asymptote); Matt Mackall (Mercurial); Richard Stallman (Emacs); and the Debian GNU/Linux project.

感谢以下机构对我在科学和数学教学工作的支持:惠特克生物医学工程基金会、赫兹基金会、剑桥大学基督圣体学院院长和研究员、麻省理工学院教学与学习实验室和本科教育院长办公室;特别是罗杰·贝克、约翰·威廉姆斯和盖茨比慈善基金会的受托人。

For supporting my work in science and mathematics teaching: The Whitaker Foundation in Biomedical Engineering; the Hertz Foundation; the Master and Fellows of Corpus Christi College, Cambridge; the MIT Teaching and Learning Laboratory and the Office of the Dean for Undergraduate Education; and especially Roger Baker, John Williams, and the Trustees of the Gatsby Charitable Foundation.

一路顺风

Bon voyage

作为我们的第一个工具,让我们欢迎一位来自物理学和工程学的访客:维度分析方法。

As our first tool, let’s welcome a visitor from physics and engineering: the method of dimensional analysis.

1

1

方面

Dimensions

我们的第一个街头格斗工具是量纲分析,简称维度。为了展示其应用的多样性,我们以一个经济学例子来介绍该工具,并结合牛顿力学和积分学的例子进行进一步的阐释。

Our first street-fighting tool is dimensional analysis or, when abbreviated, dimensions. To show its diversity of application, the tool is introduced with an economics example and sharpened on examples from Newtonian mechanics and integral calculus.

1.1 经济学:跨国公司的力量

1.1 Economics: The power of multinational corporations

全球化的批评者经常做出如下比较[ 25 ]来证明跨国公司的权力过大:

Critics of globalization often make the following comparison [25] to prove the excessive power of multinational corporations:

尼日利亚是一个经济相对强劲的国家,其GDP(国内生产总值)为990亿美元。埃克森美孚的净资产为1190亿美元。“当跨国公司的净资产高于其经营所在国的GDP时,我们谈论的究竟是什么样的权力关系?”劳拉·莫罗西尼问道。

In Nigeria, a relatively economically strong country, the GDP [gross domestic product] is $99 billion. The net worth of Exxon is $119 billion. “When multinationals have a net worth higher than the GDP of the country in which they operate, what kind of power relationship are we talking about?” asks Laura Morosini.

在继续之前,请先探讨以下问题:

Before continuing, explore the following question:

图像 埃克森美孚和尼日利亚之间的比较最严重的错误是什么?

What is the most egregious fault in the comparison between Exxon and Nigeria?

这个领域竞争激烈,但有一个缺陷显而易见。在解读GDP的含义后,这一点就显而易见了。GDP 990亿美元,其实就是每年990亿美元的货币流量。一年,也就是地球绕太阳公转一周的时间,是一种天文现象,被任意选择来衡量一种社会现象——货币流动。

The field is competitive, but one fault stands out. It becomes evident after unpacking the meaning of GDP. A GDP of $99 billion is shorthand for a monetary flow of $99 billion per year. A year, which is the time for the earth to travel around the sun, is an astronomical phenomenon that has been arbitrarily chosen for measuring a social phenomenon—namely, monetary flow.

假设经济学家选择十年作为衡量GDP的时间单位。那么尼日利亚的GDP(假设逐年保持稳定)每十年大约为1万亿美元,报告也为1万亿美元。现在,尼日利亚的GDP已经远远超过埃克森美孚,而埃克森美孚的资产仅为尼日利亚GDP的十分之一。为了得出相反的结论,假设以周作为衡量GDP的时间单位。尼日利亚的GDP变成了每周20亿美元,报告也为20亿美元。现在,孱弱的尼日利亚在强大的埃克森美孚面前束手无策,后者的规模是尼日利亚的50倍。

Suppose instead that economists had chosen the decade as the unit of time for measuring GDP. Then Nigeria’s GDP (assuming the flow remains steady from year to year) would be roughly $1 trillion per decade and be reported as $1 trillion. Now Nigeria towers over Exxon, whose puny assets are a mere one-tenth of Nigeria’s GDP. To deduce the opposite conclusion, suppose the week were the unit of time for measuring GDP. Nigeria’s GDP becomes $2 billion per week, reported as $2 billion. Now puny Nigeria stands helpless before the mighty Exxon, 50-fold larger than Nigeria.

有效的经济学论证不可能得出依赖于用于测量时间的天文现象的结论。其错误在于比较无法比较的量。净值是一个数量:它具有货币量纲,通常以美元为单位。然而,GDP是一个流量或速率:它具有单位时间的货币量纲,通常以美元/年为单位。(量纲是通用的,独立于测量体系,而单位则是该量纲在特定体系中的测量方式。)将净值与GDP进行比较,无异于将货币量与货币流量进行比较。由于它们的量纲不同,这种比较是一种范畴错误[ 39 ],因此必然会产生无意义的结果。

A valid economic argument cannot reach a conclusion that depends on the astronomical phenomenon chosen to measure time. The mistake lies in comparing incomparable quantities. Net worth is an amount: It has dimensions of money and is typically measured in units of dollars. GDP, however, is a flow or rate: It has dimensions of money per time and typical units of dollars per year. (A dimension is general and independent of the system of measurement, whereas the unit is how that dimension is measured in a particular system.) Comparing net worth to GDP compares a monetary amount to a monetary flow. Because their dimensions differ, the comparison is a category mistake [39] and is therefore guaranteed to generate nonsense.

问题 1.1 单位还是尺寸?

Problem 1.1 Units or dimensions?

米、千克和秒是单位还是量纲?能量、电荷、功率和力呢?

Are meters, kilograms, and seconds units or dimensions? What about energy, charge, power, and force?

同样有缺陷的比较是单位时间长度(速度)与长度的比较:“我走1.5米/秒——比300米高的纽约帝国大厦小得多。” 这完全是胡扯。为了得出相反但仍然胡扯的结论,可以用小时来衡量时间:“我走5400米/小时——比300米高的帝国大厦大得多。”

A similarly flawed comparison is length per time (speed) versus length: “I walk 1.5 ms−1—much smaller than the Empire State building in New York, which is 300 m high.” It is nonsense. To produce the opposite but still nonsense conclusion, measure time in hours: “I walk 5400 m/hr—much larger than the Empire State building, which is 300 m high.”

我经常看到类似尼日利亚和埃克森美孚的例子,将企业实力与国家实力进行比较。我曾写信给一位作者,解释说我理解他的结论,但他的论点包含一个致命的维度错误。他回复说,我的观点很有意思,但用数字来比较国家实力,比他写的更有说服力,所以他保留了原样!

I often see comparisons of corporate and national power similar to our Nigeria–Exxon example. I once wrote to one author explaining that I sympathized with his conclusion but that his argument contained a fatal dimensional mistake. He replied that I had made an interesting point but that the numerical comparison showing the country’s weakness was stronger as he had written it, so he was leaving it unchanged!

一个维度上有效的比较是同类比较:要么将尼日利亚的GDP与埃克森美孚的收入进行比较,要么将埃克森美孚的净资产与尼日利亚的净资产进行比较。由于各国的净资产通常不以表格形式列出,而企业收入却很容易获得,因此可以尝试将埃克森美孚的年收入与尼日利亚的GDP进行比较。到2006年,埃克森美孚已成为埃克森美孚,年收入约为3500亿美元——几乎是尼日利亚2006年GDP(2000亿美元)的两倍。这种有效的比较比有缺陷的比较更有说服力,因此保留有缺陷的比较甚至不合时宜!

A dimensionally valid comparison would compare like with like: either Nigeria’s GDP with Exxon’s revenues, or Exxon’s net worth with Nigeria’s net worth. Because net worths of countries are not often tabulated, whereas corporate revenues are widely available, try comparing Exxon’s annual revenues with Nigeria’s GDP. By 2006, Exxon had become Exxon Mobil with annual revenues of roughly $350 billion—almost twice Nigeria’s 2006 GDP of $200 billion. This valid comparison is stronger than the flawed one, so retaining the flawed comparison was not even expedient!

被比较的量必须具有相同的量纲,这是进行有效比较的必要条件,但这还不够。一个代价高昂的例子是1999年的火星气候探测器(MCO),它撞上了火星表面,而不是滑入火星轨道。根据事故调查委员会(MIB)的说法,事故原因是英制和公制单位不匹配[ 26第6页]:

That compared quantities must have identical dimensions is a necessary condition for making valid comparisons, but it is not sufficient. A costly illustration is the 1999 Mars Climate Orbiter (MCO), which crashed into the surface of Mars rather than slipping into orbit around it. The cause, according to the Mishap Investigation Board (MIB), was a mismatch between English and metric units [26, p. 6]:

MCO MIB 已确定 MCO 航天器坠毁的根本原因是未在用于轨迹模型的地面软件文件“小力”(Small Forces)的编码中使用公制单位。具体而言,在名为 SM_FORCES(小力)的软件应用程序代码中,使用的推进器性能数据采用英制单位而非公制单位。一个名为“角动量去饱和”(AMD)的文件包含 SM_FORCES 软件的输出数据。根据现有的软件接口文档,AMD 文件中的数据必须采用公制单位,而轨迹建模人员也假定数据符合要求,采用公制单位。

The MCO MIB has determined that the root cause for the loss of the MCO spacecraft was the failure to use metric units in the coding of a ground software file, Small Forces, used in trajectory models. Specifically, thruster performance data in English units instead of metric units was used in the software application code titled SM_FORCES (small forces). A file called Angular Momentum Desaturation (AMD) contained the output data from the SM_FORCES software. The data in the AMD file was required to be in metric units per existing software interface documentation, and the trajectory modelers assumed the data was provided in metric units per the requirements.

请务必注意您的尺寸和单位。

Make sure to mind your dimensions and units.

问题 1.2 寻找错误的比较

Problem 1.2 Finding bad comparisons

寻找日常的比较——例如,在新闻、报纸或互联网上——这些比较在维度上是有缺陷的。

Look for everyday comparisons—for example, on the news, in the newspaper, or on the Internet—that are dimensionally faulty.

1.2 牛顿力学:自由落体

1.2 Newtonian mechanics: Free fall

维度不仅有助于推翻错误的论证,也有助于生成正确的论证。为此,问题中的量需要具有维度。以下是一个相反的例子,说明了哪些做法不该做:许多微积分教科书是这样引入一个经典问题的:

Dimensions are useful not just to debunk incorrect arguments but also to generate correct ones. To do so, the quantities in a problem need to have dimensions. As a contrary example showing what not to do, here is how many calculus textbooks introduce a classic problem in motion:

一个静止的球从 h英尺的高度落下,以 ν英尺/秒的速度撞击地面。假设重力加速度为 g英尺/秒²,忽略空气阻力,求 ν。

A ball initially at rest falls from a height of h feet and hits the ground at a speed of ν feet per second. Find ν assuming a gravitational acceleration of g feet per second squared and neglecting air resistance.

英尺或英尺每秒等单位以粗体突出显示,因为它们出现的频率太高,否则很容易被忽略,而它们的出现会造成严重的问题。由于高度为 h 英尺,变量 h 不包含高度单位:因此 h 是无量纲的。(如果 h 有量纲,那么问题就应该简单地表述为球从高度 h 落下;那么长度的量纲就属于 h。)类似地明确指定单位意味着变量 g 和 v 也是无量纲的。由于 g、h 和 ν 是无量纲的,因此将 ν 与由 g 和 h 导出的量进行比较,都是无量纲量之间的比较。因此,它在量纲上始终有效,量纲分析无法帮助我们猜测撞击速度。

The units such as feet or feet per second are highlighted in boldface because their inclusion is so frequent as to otherwise escape notice, and their inclusion creates a significant problem. Because the height is h feet, the variable h does not contain the units of height: h is therefore dimensionless. (For h to have dimensions, the problem would instead state simply that the ball falls from a height h; then the dimension of length would belong to h.) A similar explicit specification of units means that the variables g and v are also dimensionless. Because g, h, and ν are dimensionless, any comparison of ν with quantities derived from g and h is a comparison between dimensionless quantities. It is therefore always dimensionally valid, so dimensional analysis cannot help us guess the impact speed.

放弃维度这个宝贵的工具,就像是徒手搏斗,因此,我们必须在初始条件下求解以下微分方程:

Giving up the valuable tool of dimensions is like fighting with one hand tied behind our back. Thereby constrained, we must instead solve the following differential equation with initial conditions:

图像

其中 y(t) 是球的高度,dy/dt 是球的速度,g 是重力加速度。

where y(t) is the ball’s height, dy/dt is the ball’s velocity, and g is the gravitational acceleration.

问题 1.3 微积分解决方案

Problem 1.3 Calculus solution

使用微积分证明自由落体微分方程 d 2 y/dt 2 = −g,其初始条件为 y(0) = h 且 dy/dt = 0(t = 0 时),有以下解:

Use calculus to show that the free-fall differential equation d2y/dt2 = −g with initial conditions y(0) = h and dy/dt = 0 at t = 0 has the following solution:

图像

图像 利用问题 1.3 中球的位置和速度的解,撞击速度是多少?

Using the solutions for the ball’s position and velocity in Problem 1.3, what is the impact speed?

当 y(t) = 0 时,球落地。因此撞击时间 t 0图像。撞击速度为 −gt 0或 − 图像。因此,撞击速度(无符号速度)为图像

When y(t) = 0, the ball meets the ground. Thus the impact time t0 is . The impact velocity is −gt0 or −. Therefore the impact speed (the unsigned velocity) is .

这种分析容易引发一些代数错误:求解 t 0时忘记开平方,或者计算撞击速度时用除法而不是乘以 g。练习——换句话说,犯下并纠正许多错误——可以降低这些错误在简单问题中的发生率,但步骤繁多的复杂问题仍然是雷区。我们希望找到更不容易出错的方法。

This analysis invites several algebra mistakes: forgetting to take a square root when solving for t0, or dividing rather than multiplying by g when finding the impact velocity. Practice—in other words, making and correcting many mistakes—reduces their prevalence in simple problems, but complex problems with many steps remain minefields. We would like less error-prone methods.

一种稳健的替代方案是量纲分析法。但这种方法要求ν、g和h中至少有一个量具有量纲。否则,所有候选撞击速度,无论多么荒谬,都等同于无量纲量,因此具有有效的量纲。

One robust alternative is the method of dimensional analysis. But this tool requires that at least one quantity among ν, g, and h have dimensions. Otherwise, every candidate impact speed, no matter how absurd, equates dimensionless quantities and therefore has valid dimensions.

因此,让我们重新表述自由落体问题,以便数量保留其维度:

Therefore, let’s restate the free-fall problem so that the quantities retain their dimensions:

一个静止的球从高度 h 落下,以速度 ν 撞击地面。假设重力加速度为 g,忽略空气阻力,求 ν。

A ball initially at rest falls from a height h and hits the ground at speed ν. Find ν assuming a gravitational acceleration g and neglecting air resistance.

首先,重述比原文措辞更简短、更清晰:

The restatement is, first, shorter and crisper than the original phrasing:

一个静止的球从 h 英尺的高度落下,以 ν 英尺/秒的速度落地。假设重力加速度为 g 英尺/秒²,忽略空气阻力,求 ν。

A ball initially at rest falls from a height of h feet and hits the ground at a speed of ν feet per second. Find ν assuming a gravitational acceleration of g feet per second squared and neglecting air resistance.

其次,重述更具通用性。它没有对单位制做出任何假设,因此即使长度单位是米、腕尺或弗隆,它仍然适用。最重要的是,重述给出了h、g和ν的量纲。它们的量纲几乎可以唯一地决定撞击速度——无需我们解微分方程。

Second, the restatement is more general. It makes no assumption about the system of units, so it is useful even if meters, cubits, or furlongs are the unit of length. Most importantly, the restatement gives dimensions to h, g, and ν. Their dimensions will almost uniquely determine the impact speed—without our needing to solve a differential equation.

高度 h 的量纲为长度,或简称为 L。重力加速度 g 的量纲为长度/时间平方,或 LT −2,其中 T 表示时间维度。速度的量纲为 LT −1,因此 ν 是 g 和 h 的函数,其量纲为 LT −1

The dimensions of height h are simply length or, for short, L. The dimensions of gravitational acceleration g are length per time squared or LT−2, where T represents the dimension of time. A speed has dimensions of LT−1, so ν is a function of g and h with dimensions of LT−1.

问题 1.4 熟悉量的维度

Problem 1.4 Dimensions of familiar quantities

就长度L、质量M、时间T这几个基本量纲而言,能量、功率、扭矩的量纲分别是什么?

In terms of the basic dimensions length L, mass M, and time T, what are the dimensions of energy, power, and torque?

图像 gh的哪种组合具有速度维度?

What combination of g and h has dimensions of speed?

这种组合图像具有速度维度。

The combination has dimensions of speed.

图像

图像 图像gh的组合是否只具有速度维度?

Is the only combination of g and h with dimensions of speed?

为了确定是否图像是唯一的可能性,可以使用约束传播[ 43 ]。最强的约束是,g和h的组合(作为速度)应该具有逆时间(T -1)的维度。由于h不包含时间维度,因此它无法帮助构造T -1。因为g 包含 T −2,则 T −1必定来自图像。第二个约束条件是 组合包含 L 1图像已经贡献了 L 1/2,因此缺失的 L 1/2必定来自图像这两个约束条件唯一地决定了 g 和 h 在撞击速度 ν 中的表现形式。

In order to decide whether is the only possibility, use constraint propagation [43]. The strongest constraint is that the combination of g and h, being a speed, should have dimensions of inverse time (T−1). Because h contains no dimensions of time, it cannot help construct T−1. Because g contains T−2, the T−1 must come from . The second constraint is that the combination contain L1. The already contributes L1/2, so the missing L1/2 must come from The two constraints thereby determine uniquely how g and h appear in the impact speed ν.

然而,ν 的确切表达式并非唯一。它可能是图像图像,或者通常为图像× 无量纲常数。乘以无量纲常数的惯用法经常出现,因此应该使用类似于等号的简洁符号:

The exact expression for ν is, however, not unique. It could be , , or, in general, × dimensionless constant. The idiom of multiplication by a dimensionless constant occurs frequently and deserves a compact notation akin to the equals sign:

图像

包括这个∼符号在内,我们有几种相等的情况:

Including this ∼ notation, we have several species of equality:

图像

精确的撞击速度为图像,因此维度结果图像包含整个函数依赖关系!它只缺少无量纲因子图像,而这些因素通常并不重要。在此示例中,高度可能从几厘米(跳蚤跳跃)到几米(猫从窗台上跳下来)不等。高度的 100 倍变化会导致撞击速度发生 10 倍变化。同样,重力加速度可能从 0.27 ms −2(在小行星谷神星上)到 25 ms −2(在木星上)不等。g 的 100 倍变化会导致撞击速度再次发生 10 倍变化。因此,撞击速度的许多变化不是来自无量纲因子图像,而是来自符号因子——这些符号因子是通过量纲分析精确计算出来的。

The exact impact speed is , so the dimensions result contains the entire functional dependence! It lacks only the dimensionless factor , and these factors are often unimportant. In this example, the height might vary from a few centimeters (a flea hopping) to a few meters (a cat jumping from a ledge). The factor-of-100 variation in height contributes a factor-of-10 variation in impact speed. Similarly, the gravitational acceleration might vary from 0.27 ms−2 (on the asteroid Ceres) to 25 ms−2 (on Jupiter). The factor-of-100 variation in g contributes another factor-of-10 variation in impact speed. Much variation in the impact speed, therefore, comes not from the dimensionless factor but rather from the symbolic factors—which are computed exactly by dimensional analysis.

此外,不计算精确答案也可能是一个优势。精确答案包含所有因数和项,这使得一些不太重要的信息(例如无量纲因子)图像能够掩盖诸如 之类的重要信息图像。正如威廉·詹姆斯所说:“智慧的艺术在于知道该忽略什么”[ 19,第22章]。

Furthermore, not calculating the exact answer can be an advantage. Exact answers have all factors and terms, permitting less important information, such as the dimensionless factor . to obscure important information such as . As William James advised, “The art of being wise is the art of knowing what to overlook” [19, Chapter 22].

问题 1.5 垂直投掷

Problem 1.5 Vertical throw

你以速度ν0垂直向上抛出一个球。利用量纲分析估算球返回你手中所需的时间(忽略空气阻力)。然后通过解自由落体微分方程得出确切时间。量纲分析结果中缺少哪个无量纲因子?

You throw a ball directly upward with speed ν0. Use dimensional analysis to estimate how long the ball takes to return to your hand (neglecting air resistance). Then find the exact time by solving the free-fall differential equation. What dimensionless factor was missing from the dimensional-analysis result?

1.3 猜测积分

1.3 Guessing integrals

自由落体运动的分析(第 1.2 节)表明,不将有量纲的量与其单位分开是有价值的。然而,如果这些量是无量纲的,例如以下高斯积分中的 5 和 x,情况又会怎样呢?

The analysis of free fall (Section 1.2) shows the value of not separating dimensioned quantities from their units. However, what if the quantities are dimensionless, such as the 5 and x in the following Gaussian integral:

图像

或者,维度可能是未指定的——这在数学中很常见,因为它是一种通用语言。例如,概率论使用高斯积分

Alternatively, the dimensions might be unspecified—a common case in mathematics because it is a universal language. For example, probability theory uses the Gaussian integral

图像

其中 x 可以是高度、探测器误差或其他参数。热物理学使用类似的积分

where x could be height, detector error, or much else. Thermal physics uses the similar integral

图像

其中ν是分子速度。数学作为通用语言,研究它们的共同形式图像,而无需指定α和x的维度。缺乏具体性赋予了数学抽象的能力,但也使得量纲分析变得困难。

where ν is a molecular speed. Mathematics, as the common language, studies their common form without specifying the dimensions of α and x. The lack of specificity gives mathematics its power of abstraction, but it makes using dimensional analysis difficult.

图像 如何应用维度分析而不失去数学抽象的好处?

How can dimensional analysis be applied without losing the benefits of mathematical abstraction?

答案是找到未指定维度的量,然后为它们分配一组一致的维度。为了说明这种方法,我们将其应用于广义定高斯积分

The answer is to find the quantities with unspecified dimensions and then to assign them a consistent set of dimensions. To illustrate the approach, let’s apply it to the general definite Gaussian integral

图像

与其特定的近亲(α = 5,即积分)不同图像,一般形式没有指定 x 或 α 的维度——并且开放性提供了使用维度分析方法所需的自由。

Unlike its specific cousin with α = 5, which is the integral , the general form does not specify the dimensions of x or α—and that openness provides the freedom needed to use the method of dimensional analysis.

该方法要求任何方程的维度有效。因此,在下列方程中,左右两边的维度必须相同:

The method requires that any equation be dimensionally valid. Thus, in the following equation, the left and right sides must have identical dimensions:

图像

图像 右边是x的函数吗?是α的函数吗?它包含积分常数吗?

Is the right side a function of x? Is it a function of α? Does it contain a constant of integration?

左边除了 x 和 α 之外不包含任何符号量。但 x 是积分变量,且积分在一定范围内,所以积分时 x 消失(并且不出现积分常数)。因此,右边——那个“某物”——只是 α 的函数。用符号表示,

The left side contains no symbolic quantities other than x and α. But x is the integration variable and the integral is over a definite range, so x disappears upon integration (and no constant of integration appears). Therefore, the right side—the “something”—is a function only of α. In symbols,

图像

函数 f 可能包含无量纲数,例如 2/3 或图像,但 α 是其唯一有量纲的输入。

The function f might include dimensionless numbers such as 2/3 or , but α is its only input with dimensions.

为了使方程在维度上有效,积分必须具有与 f(α) 相同的维度,并且 f(α) 的维度取决于 α 的维度。因此,维度分析程序包含以下三个步骤:

For the equation to be dimensionally valid, the integral must have the same dimensions as f(α), and the dimensions of f(α) depend on the dimensions of α. Accordingly, the dimensional-analysis procedure has the following three steps:

步骤 1. 为 α 分配尺寸(第 1.3.1 节)。

Step 1. Assign dimensions to α (Section 1.3.1).

步骤 2. 找到积分的维度(第 1.3.2 节)。

Step 2. Find the dimensions of the integral (Section 1.3.2).

步骤 3. 使用这些维度制作 f(α)(第 1.3.3 节)。

Step 3. Make an f(α) with those dimensions (Section 1.3.3).

1.3.1 为α分配维度

1.3.1 Assigning dimensions to α

参数 α 出现在指数中。指数指定将一个量乘以自身的次数。例如,这里是 2 n

The parameter α appears in an exponent. An exponent specifies how many times to multiply a quantity by itself. For example, here is 2n:

图像

“多少次”的概念是一个纯数字,因此指数是无量纲的。

The notion of “how many times” is a pure number, so an exponent is dimensionless.

因此,高斯积分中的指数 —αx 2是无量纲的。为方便起见,α 的维度记为 [α],x 的维度记为 [x]。则

Hence the exponent —αx2 in the Gaussian integral is dimensionless. For convenience, denote the dimensions of α by [α] and of x by [x]. Then

图像

或者

or

图像

这个结论很有用,但继续使用未指定但通用的维度需要大量的符号,而这些符号可能会掩盖推理。

This conclusion is useful, but continuing to use unspecified but general dimensions requires lots of notation, and the notation risks burying the reasoning.

最简单的替代方案是让 x 无量纲化。这样一来,α 和 f(α) 也就无量纲化了,因此 f(α) 的任何候选值在量纲上都是有效的,这样一来量纲分析就毫无意义了。最简单有效的替代方案是赋予 x 简单的量纲,例如长度。(如果你把 x 轴想象成平放在地板上,那么这个选择就很自然了。)那么 [α] = L −2

The simplest alternative is to make x dimensionless. That choice makes α and f(α) dimensionless, so any candidate for f(α) would be dimensionally valid, making dimensional analysis again useless. The simplest effective alternative is to give x simple dimensions—for example, length. (This choice is natural if you imagine the x axis lying on the floor.) Then [α] = L−2.

1.3.2 积分的维度

1.3.2 Dimensions of the integral

赋值 [x] = L 和 [α] = L −2决定了高斯积分的维数。以下是该积分:

The assignments [x] = L and [α] = L−2 determine the dimensions of the Gaussian integral. Here is the integral again:

图像

积分的维数取决于其三个部分的维数:积分符号 ∫、被积函数 e −αx 2和微分 dx。积分符号源于德语单词Summe的拉长字母 S ,Summe 意为“和”。在有效的和式中,所有项都具有相同的维数:维数的基本原理要求“苹果只能加在苹果上”。出于同样的原因,整个和式的维数与任何项都相同。因此,求和符号(以及积分符号)不影响维数:积分符号是无维数的。

The dimensions of an integral depend on the dimensions of its three pieces: the integral sign ∫, the integrand e−αx2, and the differential dx. The integral sign originated as an elongated S for Summe, the German word for sum. In a valid sum, all terms have identical dimensions: The fundamental principle of dimensions requires that apples be added only to apples. For the same reason, the entire sum has the same dimensions as any term. Thus, the summation sign—and therefore the integration sign—do not affect dimensions: The integral sign is dimensionless.

问题 1.6 积分速度

Problem 1.6 Integrating velocity

位置是速度的积分。然而,位置和速度的量纲不同。这种差异如何与积分符号无量纲的结论相一致?

Position is the integral of velocity. However, position and velocity have different dimensions. How is this difference consistent with the conclusion that the integration sign is dimensionless?

由于积分符号无量纲,因此积分的量纲等于指数因子 e −αx 2的量纲乘以 dx 的量纲。指数函数尽管有 −αx 2这样的大指数,但实际上只是几个 e 的乘积。由于 e 是无量纲的,因此 e −αx 2也是无量纲的。

Because the integration sign is dimensionless, the dimensions of the integral are the dimensions of the exponential factor e−αx2 multiplied by the dimensions of dx. The exponential, despite its fierce exponent −αx2, is merely several copies of e multiplied together. Because e is dimensionless, so is e−αx2.

图像 dx的尺寸是多少

What are the dimensions of dx?

要计算 dx 的维数,请遵循 Silvanus Thompson [ 45 , p. 1 ]的建议:将 d 读作“一点点”。那么 dx 就是“x 的一点点”。一点点长度仍然是长度,所以 dx 是长度。通常,dx 的维数与 x 相同。同样,d(∫ 的倒数)是无维数的。

To find the dimensions of dx, follow the advice of Silvanus Thompson [45, p. 1]: Read d as “a little bit of.” Then dx is “a little bit of x.” A little length is still a length, so dx is a length. In general, dx has the same dimensions as x. Equivalently, d—the inverse of ∫—is dimensionless.

组装各个部分,整个积分的长度尺寸为:

Assembling the pieces, the whole integral has dimensions of length:

图像

问题 1.7 积分不计算面积吗?

Problem 1.7 Don’t integrals compute areas?

人们普遍认为积分计算的是面积。面积的维数是 L ²。那么高斯积分的维数怎么会是 L 呢?

A common belief is that integration computes areas. Areas have dimensions of L2. How then can the Gaussian integral have dimensions of L?

1.3.3 制作具有正确维度的f(α)

1.3.3 Making an f(α) with correct dimensions

量纲分析的第三步也是最后一步是构造一个与积分维度相同的 f(α)。由于 α 的维度为 L −2,因此将 α 转化为长度的唯一方法是构造 α −1/2。因此,

The third and final step in this dimensional analysis is to construct an f(α) with the same dimensions as the integral. Because the dimensions of α are L−2, the only way to turn α into a length is to form α−1/2. Therefore,

图像

这个有用的结果仅缺少一个无量纲因子,并且无需任何积分即可获得。

This useful result, which lacks only a dimensionless factor, was obtained without any integration.

为了确定无量纲常数,设 α = 1 并评估

To determine the dimensionless constant, set α = 1 and evaluate

图像

这个经典积分将在2.1 节中近似,猜测为图像。两个结果 f(1) =图像和 f(α) ∼ α −1/2要求 f(α) = 图像,从而得出

This classic integral will be approximated in Section 2.1 and guessed to be . The two results f(1) = and f(α) ∼ α−1/2 require that f(α) = , which yields

图像

我们经常记住无量纲常数,却忘记了α的幂。千万不要这样做。α因子通常比无量纲常数重要得多。方便的是,α因子是量纲分析可以计算出来的。

We often memorize the dimensionless constant but forget the power of α. Do not do that. The α factor is usually much more important than the dimensionless constant. Conveniently, the α factor is what dimensional analysis can compute.

问题 1.8 变量的变化

Problem 1.8 Change of variable

回到第8页,假设你不知道f(α)。不进行量纲分析,证明图像

Rewind back to page 8 and pretend that you do not know f(α). Without doing dimensional analysis, show that .

问题 1.9 简单情况α = 1

Problem 1.9 Easy case α = 1

设 α = 1 是简单推理的一个例子(第二章),它违反了 x 为长度且 α 的维度为 L −2的假设。为什么设 α = 1 是可以的?

Setting α = 1, which is an example of easy-cases reasoning (Chapter 2), violates the assumption that x is a length and α has dimensions of L−2. Why is it okay to set α = 1?

问题 1.10 积分困难的指数

Problem 1.10 Integrating a difficult exponential

运用量纲分析法进行调查图像

Use dimensional analysis to investigate .

1.4 总结和进一步的问题

1.4 Summary and further problems

不要把苹果和橘子混为一谈:等式或和式中的每个项必须具有相同的量纲!这个限制是一个强大的工具。它帮助我们在不积分的情况下求积分,并预测微分方程的解。这里有更多练习这个工具的问题。

Do not add apples to oranges: Every term in an equation or sum must have identical dimensions! This restriction is a powerful tool. It helps us to evaluate integrals without integrating and to predict the solutions of differential equations. Here are further problems to practice this tool.

问题 1.11 使用维度的积分

Problem 1.11 Integrals using dimensions

使用维度分析来查找图像一个图像有用的结果是

Use dimensional analysis to find and A useful result is

图像

问题 1.12 斯特藩-玻尔兹曼定律

Problem 1.12 Stefan–Boltzmann law

黑体辐射是一种电磁现象,因此辐射强度取决于光速 c。它也是一种热现象,因此它取决于热能 k B T,其中 T 是物体的温度,k B是玻尔兹曼常数。它是一种量子现象,因此它取决于普朗克常数图像。因此,黑体辐射强度 I 取决于 c、k B T 和图像。利用量纲分析证明 I 图像T 4并求出比例常数 σ。然后查找缺失的无量纲常数。(这些结果将在第 5.3.3 节中使用。)

Blackbody radiation is an electromagnetic phenomenon, so the radiation intensity depends on the speed of light c. It is also a thermal phenomenon, so it depends on the thermal energy kBT, where T is the object’s temperature and kB is Boltzmann’s constant. And it is a quantum phenomenon, so it depends on Planck’s constant . Thus the blackbody-radiation intensity I depends on c, kBT, and . Use dimensional analysis to show that I T4 and to find the constant of proportionality σ. Then look up the missing dimensionless constant. (These results are used in Section 5.3.3.)

问题 1.13 反正弦积分

Problem 1.13 Arcsine integral

使用维度分析来找到图像一个有用的结果是

Use dimensional analysis to find A useful result is

图像

问题 1.14 相关利率

Problem 1.14 Related rates

将水以 dV/dt = 10 m 3 s −1的速率倒入一个大的倒锥体(开口角为 90°)。当水深 h = 5 m 时,估算深度增加的速率。然后使用微积分计算出准确的速率。

Water is poured into a large inverted cone (with a 90° opening angle) at a rate dV/dt = 10 m3 s−1. When the water depth is h = 5 m, estimate the rate at which the depth is increasing. Then use calculus to find the exact rate.

图像

问题 1.15 开普勒第三定律

Problem 1.15 Kepler’s third law

牛顿万有引力定律(著名的平方反比定律)指出,两个物体之间的引力是

Newton’s law of universal gravitation—the famous inverse-square law—says that the gravitational force between two masses is

图像

其中 G 是牛顿常数,m 1和 m 2是两个质量,r 是它们之间的距离。对于绕太阳运行的行星,万有引力与牛顿第二定律相结合,可得出

where G is Newton’s constant, m1 and m2 are the two masses, and r is their separation. For a planet orbiting the sun, universal gravitation together with Newton’s second law gives

图像

其中M为太阳质量,m为行星质量,r为太阳到行星的矢量,为r图像方向的单位矢量。

where M is the mass of the sun, m the mass of the planet, r is the vector from the sun to the planet, and is the unit vector in the r direction.

轨道周期τ与轨道半径r有何关系?查阅开普勒第三定律,并将你的结果与之进行比较。

How does the orbital period τ depend on orbital radius r? Look up Kepler’s third law and compare your result to it.

2

2

简单案例

Easy cases

正确的解法适用于所有情况,包括简单情况。这条准则构成了第二个工具——简单情况法的基础。它将帮助我们猜测积分、推导体积,并求解复杂的微分方程。

A correct solution works in all cases, including the easy ones. This maxim underlies the second tool—the method of easy cases. It will help us guess integrals, deduce volumes, and solve exacting differential equations.

2.1 重温高斯积分

2.1 Gaussian integral revisited

作为第一个应用,让我们重新回顾一下第 1.3 节中的高斯积分,

As the first application, let’s revisit the Gaussian integral from Section 1.3,

图像

图像 是积分图像还是图像

Is the integral or ?

正确的选择必须适用于所有图像。在此范围的端点 处图像,积分很容易计算。

The correct choice must work for all . At this range’s endpoints , the integral is easy to evaluate.

图像 图像

What is the integral when

图像

像第一种简单情况一样,将 α 增加到 ∞ 时,即使 x 很小,− αx 2也会变得非常负。一个大的负数的指数很小,因此钟形曲线会变窄成一条细长条,其面积也会缩小到零。因此,当 α → ∞ 时,积分会缩小到零。这个结果反驳了以下选项图像,当 α = ∞ 时为无穷大;并且它支持选项图像,当 α = ∞ 时为零。

As the first easy case, increase α to ∞ Then − αx2 becomes very negative, even when x is tiny. The exponential of a large negative number is tiny, so the bell curve narrows to a sliver, and its area shrinks to zero. Therefore, as α → ∞ the integral shrinks to zero. This result refutes the option , which is infinite when α = ∞; and it supports the option , which is zero when α = ∞.

图像 当α = 0时积分是多少

What is the integral when α = 0?

在 α = 0 的极端情况下,钟形曲线会变平为一条高度为单位的水平线。其面积在无穷大范围内积分后为无穷大。这一结果反驳了图像α = 0 时为零的选项;而支持了图像α = 0 时为无穷大的选项。因此,该图像选项既不通过两个简单情况测试,又图像通过了两个简单情况测试。

In the α = 0 extreme, the bell curve flattens into a horizontal line with unit height. Its area, integrated over the infinite range, is infinite. This result refutes the option, which is zero when α = 0; and it supports the option, which is infinity when α = 0. Thus the option fails both easy-cases tests, and the option passes both easy-cases tests.

图像

如果只有这两个选项,我们会选择图像。但是,如果第三个选项是图像,你该如何在它和 之间做出选择图像?这两个选项都通过了简单情况测试;它们也具有相同的维度。选择看起来很困难。

If these two options were the only options, we would choose . However, if a third option were , how could you decide between it and ? Both options pass both easy-cases tests; they also have identical dimensions. The choice looks difficult.

要选择,请尝试第三种简单情况:α = 1。然后积分简化为

To choose, try a third easy case: α = 1. Then the integral simplifies to

图像

这个经典积分可以用极坐标来求出闭式,但这种方法也需要一些技巧,而且应用范围有限(多元微积分的教科书会给出详细的说明)。一种不太优雅但更通用的方法是先数值计算积分,然后利用近似值来猜测闭式。

This classic integral can be evaluated in closed form by using polar coordinates, but that method also requires a trick with few other applications (textbooks on multivariable calculus give the gory details). A less elegant but more general approach is to evaluate the integral numerically and to use the approximate value to guess the closed form.

图像

因此,将光滑曲线 e −x 2替换为包含 n 条线段的曲线。这种分段线性近似将面积转化为 n 个梯形的和。当 n 趋近于无穷大时,梯形的面积越来越接近光滑曲线下方的面积。

Therefore, replace the smooth curve e−x2 with a curve having n line segments. This piecewise-linear approximation turns the area into a sum of n trapezoids. As n approaches infinity, the area of the trapezoids more and more closely approaches the area under the smooth curve.

该表给出了将曲线分成 n 条线段后,在 x = −10 ... 10 范围内的曲线下面积。面积最终稳定在一个值上,看起来很熟悉。它从 1.7 开始,这可能源于图像。然而,它继续下降到 1.77,这太大了,不可能是图像。幸运的是,π 略大于 3,因此面积可能收敛到图像

The table gives the area under the curve in the range x = −10 . . . 10, after dividing the curve into n line segments. The areas settle onto a stable value, and it looks familiar. It begins with 1.7, which might arise from . However, it continues as 1.77, which is too large to be . Fortunately, π is slightly larger than 3, so the area might be converging to .

图像

让我们通过将平方面积与π进行比较来检查:

Let’s check by comparing the squared area against π:

图像

接近的匹配表明 α = 1 高斯积分确实是图像

The close match suggests that the α = 1 Gaussian integral is indeed

图像

因此,一般的高斯积分

Therefore the general Gaussian integral

图像

当 α = 1 时,必须减少到。图像它也必须在另外两个简单情况 α = 0 和 α = ∞ 下正确运行。

must reduce to when α = 1. It must also behave correctly in the other two easy cases α = 0 and α = ∞.

在三个选项中图像图像只有α=0、1和∞三个检验通过。因此图像图像

Among the three choices , , and , only passes all three tests α = 0, 1, and ∞. Therefore,

图像

简单案例并非判断这些选择的唯一方法。例如,量纲分析也可以限制可能性(第 1.3 节)。它甚至会排除像图像通过所有三个简单案例测试那样的选择。然而,简单案例的设计本身就很简单。它们不需要我们为 x、α 和 dx 创造或推导维度(第 1.3 节的扩展分析)。与量纲分析不同,简单案例也可以排除像图像具有正确维度那样的选择。每种工具都有其优势。

Easy cases are not the only way to judge these choices. Dimensional analysis, for example, can also restrict the possibilities (Section 1.3). It even eliminates choices like that pass all three easy-cases tests. However, easy cases are, by design, simple. They do not require us to invent or deduce dimensions for x, α, and dx (the extensive analysis of Section 1.3). Easy cases, unlike dimensional analysis, can also eliminate choices like with correct dimensions. Each tool has its strengths.

问题 2.1 测试几种替代方案

Problem 2.1 Testing several alternatives

对于高斯积分

For the Gaussian integral

图像

使用三个简单案例测试来评估以下候选方案的价值。

use the three easy-cases tests to evaluate the following candidates for its value.

图像

问题 2.2 合理的、不正确的替代方案

Problem 2.2 Plausible, incorrect alternative

图像是否有具有有效尺寸并通过三个简单案例测试的替代方案?

Is there an alternative to that has valid dimensions and passes the three easy-cases tests?

问题 2.3 猜测封闭形式

Problem 2.3 Guessing a closed form

使用变量变换来表明

Use a change of variable to show that

图像

第二个积分的积分范围有限,因此比第一个积分更容易进行数值计算。使用梯形近似法和计算机或可编程计算器估算第二个积分。然后猜测第一个积分的封闭形式。

The second integral has a finite integration range, so it is easier than the first integral to evaluate numerically. Estimate the second integral using the trapezoid approximation and a computer or programmable calculator. Then guess a closed form for the first integral.

2.2 平面几何:椭圆的面积

2.2 Plane geometry: The area of an ellipse

第二个简单应用来自平面几何:椭圆的面积。这个椭圆的长半轴为 a,短半轴为 b。对于它的面积 A,考虑以下候选:

The second application of easy cases is from plane geometry: the area of an ellipse. This ellipse has semimajor axis a and semiminor axis b. For its area A consider the following candidates:

图像

图像

图像 每位候选人的优点和缺点是什么?

What are the merits or drawbacks of each candidate?

候选 A = ab 2 的维数为 L 3,而面积的维数必定为 L 2。因此 ab 2必定是错误的。

The candidate A = ab2 has dimensions of L3, whereas an area must have dimensions of L2. Thus ab2 must be wrong.

候选 A = a 2 + b 2具有正确的尺寸(其余候选也一样),因此接下来的测试是关于半径 a 和 b 的简单情况。对于 a,最小极值 a = 0 会产生一个面积为零的无限细椭圆。然而,当 a = 0 时,候选 A = a 2 + b 2会简化为 A = b 2而不是 0;因此 a 2 + b 2未通过 a = 0 测试。

The candidate A = a2 + b2 has correct dimensions (as do the remaining candidates), so the next tests are the easy cases of the radii a and b. For a, the low extreme a = 0 produces an infinitesimally thin ellipse with zero area. However, when a = 0 the candidate A = a2 + b2 reduces to A = b2 rather than to 0; so a2 + b2 fails the a = 0 test.

当 a = 0 时,候选 A = a 3 /b 正确地预测了面积为零。由于 a = 0 是一个有用的简单情况,并且轴标签 a 和 b 几乎可以互换,因此其对称对应项 b = 0 也应该是一个有用的简单情况。它也会产生一个面积为零的无限细椭圆;可惜的是,候选 a 3 /b 预测的面积为无限大,因此它无法通过 b = 0 的检验。剩下两个候选。

The candidate A = a3/b correctly predicts zero area when a = 0. Because a = 0 was a useful easy case, and the axis labels a and b are almost interchangeable, its symmetric counterpart b = 0 should also be a useful easy case. It too produces an infinitesimally thin ellipse with zero area; alas, the candidate a3/b predicts an infinite area, so it fails the b = 0 test. Two candidates remain.

候选 A = 2ab 看起来不错。当 a = 0 或 b = 0 时,实际面积和预测面积均为零,因此 A = 2ab 通过了两种简单情况的测试。进一步的测试需要第三种简单情况:a = b。此时椭圆变成了半径为 a 、面积为 πa² 的圆然而,候选 2ab 会简化为 A = 2a² 因此它未能通过 a = b 测试。

The candidate A = 2ab shows promise. When a = 0 or b = 0, the actual and predicted areas are zero, so A = 2ab passes both easy-cases tests. Further testing requires the third easy case: a = b. Then the ellipse becomes a circle with radius a and area πa2. The candidate 2ab, however, reduces to A = 2a2, so it fails the a = b test.

候选区域 A = πab 通过了所有三个测试:a = 0、b = 0 和 a = b。随着每次测试通过,我们对候选区域的信心不断增强;πab 确实是正确的区域(问题 2.4)。

The candidate A = πab passes all three tests: a = 0, b = 0, and a = b. With each passing test, our confidence in the candidate increases; and πab is indeed the correct area (Problem 2.4).

问题 2.4 用微积分计算面积

Problem 2.4 Area by calculus

使用积分来证明 A = πab。

Use integration to show that A = πab.

问题 2.5 发明一个合格的候选人

Problem 2.5 Inventing a passing candidate

您能否为该区域发明第二个候选区域,使其具有正确的尺寸并通过 a = 0、b = 0 和 a = b 测试?

Can you invent a second candidate for the area that has correct dimensions and passes the a = 0, b = 0, and a = b tests?

问题 2.6 泛化

Problem 2.6 Generalization

猜测主半径为 a、b 和 c 的椭圆体的体积。

Guess the volume of an ellipsoid with principal radii a, b, and c.

2.3 立体几何:截头金字塔的体积

2.3 Solid geometry: The volume of a truncated pyramid

高斯积分示例(第 2.1 节)和椭圆面积示例(第 2.2 节)展示了简单案例作为一种分析方法:用于检查公式是否正确。下一个复杂层次是将简单案例作为一种综合方法:用于构造公式。

The Gaussian-integral example (Section 2.1) and the ellipse-area example (Section 2.2) showed easy cases as a method of analysis: for checking whether formulas are correct. The next level of sophistication is to use easy cases as a method of synthesis: for constructing formulas.

举个例子,取一个方形底座的金字塔,用一把与底座平行的刀从其顶部切下一块。这个截头金字塔(称为平截头体)有一个方形底座和一个与底座平行的方形顶部。设h为其垂直高度,b为其底座边长,a为其顶部边长。

As an example, take a pyramid with a square base and slice a piece from its top using a knife parallel to the base. This truncated pyramid (called the frustum) has a square base and square top parallel to the base. Let h be its vertical height, b be the side length of its base, and a be the side length of its top.

图像

图像 截头金字塔的体积是多少?

What is the volume of the truncated pyramid?

让我们综合一下体积公式。它是三个长度 h、a 和 b 的函数。这些长度分为两类:高和底长。例如,将立体倒置会交换 a 和 b 的含义,但保留 h;并且没有任何简单的运算可以将高与 a 或 b 互换。因此,体积可能有两个因子,每个因子只包含一种或多种长度:

Let’s synthesize the formula for the volume. It is a function of the three lengths h, a, and b. These lengths split into two kinds: height and base lengths. For example, flipping the solid on its head interchanges the meanings of a and b but preserves h; and no simple operation interchanges height with a or b. Thus the volume probably has two factors, each containing a length or lengths of only one kind:

图像

比例推理将确定 f;一点维度推理和大量简单案例推理将确定 g。

Proportional reasoning will determine f; a bit of dimensional reasoning and a lot of easy-cases reasoning will determine g.

图像 f是多少:体积应如何取决于高度?

What is f : How should the volume depend on the height?

要计算 f,可以使用比例推理思维实验。将固体切成垂直的细条,每条都像一个石油钻芯;然后想象将 h 加倍。这种变化使每条细条的体积加倍,从而使总体积 V 加倍。因此 f ∼ h 且 V ∝ h:

To find f, use a proportional-reasoning thought experiment. Chop the solid into vertical slivers, each like an oil-drilling core; then imagine doubling h. This change doubles the volume of each sliver and therefore doubles the whole volume V. Thus f ∼ h and V ∝ h:

图像

图像

图像 g是多少:体积应如何取决于ab

What is g : How should the volume depend on a and b?

由于 V 的维数为 L 3,函数 g(a, b) 的维数为 L 2。该约束条件是维数分析所能给出的全部内容。为了合成 g ,还需要进一步的约束条件,这些约束条件可以通过简单案例法来提供。

Because V has dimensions of L3, the function g(a, b) has dimensions of L2. That constraint is all that dimensional analysis can say. Further constraints are needed to synthesize g, and these constraints are provided by the method of easy cases.

2.3.1 简单案例

2.3.1 Easy cases

图像 ab的简单情况有哪些

What are the easy cases of a and b?

最简单的情况是极端情况 a = 0(普通金字塔)。a 和 b 之间的对称性表明还有另外两种简单情况,即 a = b 和极端情况 b = 0。简单情况有三种:

The easiest case is the extreme case a = 0 (an ordinary pyramid). The symmetry between a and b suggests two further easy cases, namely a = b and the extreme case b = 0. The easy cases are then threefold:

图像

当 a = 0 时,该立体是一个普通的金字塔,g 仅是底边长 b 的函数。由于 g 的维数为 L ²,因此 g 的唯一可能性是 g ∼ b ²;此外,V ∝ h;所以,V ∼ hb ²。当 b = 0 时,该立体是一个 b = 0 金字塔的倒置版本,因此其体积为 V ∼ ha ²。当 a = b 时,该立体是一个长方体,其体积为 V = ha ²(或 hb ²)。

When a = 0, the solid is an ordinary pyramid, and g is a function only of the base side length b. Because g has dimensions of L2, the only possibility for g is g ∼ b2; in addition, V ∝ h; so, V ∼ hb2. When b = 0, the solid is an upside-down version of the b = 0 pyramid and therefore has volume V ∼ ha2. When a = b, the solid is a rectangular prism having volume V = ha2 (or hb2).

图像 是否存在满足三个简单情况约束的体积公式?

Is there a volume formula that satisfies the three easy-cases constraints?

a = 0 和 b = 0 的约束条件由对称和 V ∼ h(a 2 + b 2 ) 满足。如果缺失的无量纲常数为 1/2,使得 V = h(a 2 + b 2 )/2,则体积也满足 a = b 的约束条件,普通金字塔(a = 0)的体积为 hb 2 /2。

The a = 0 and b = 0 constraints are satisfied by the symmetric sum V ∼ h(a2 + b2). If the missing dimensionless constant is 1/2, making V = h(a2 + b2)/2, then the volume also satisfies the a = b constraint, and the volume of an ordinary pyramid (a = 0) would be hb2/2.

图像 a = 0 时,预测V = hb 2 /2正确吗?

When a = 0, is the prediction V = hb2/2 correct?

检验预测需要找到 V ∼ hb 2中的精确无量纲常数。这项任务看起来像一道微积分题:将金字塔切成薄片,然后将它们的体积相加(积分)。不过,一个简单的替代方法是再次应用简单的案例。

Testing the prediction requires finding the exact dimensionless constant in V ∼ hb2. This task looks like a calculus problem: Slice a pyramid into thin horizontal sections and add (integrate) their volumes. However, a simple alternative is to apply easy cases again.

图像

在我们解决了一个类似但更简单的问题之后,这个简单的情况就更容易构建了:求底边为 b、高为 h 的三角形的面积。面积满足 A ∼ hb,但无量纲常数是多少?要求它,选择 b 和 h 来构成一个简单的三角形:一个 h = b 的直角三角形。两个这样的三角形可以构成一个简单的形状:一个面积为 b² 的正方形因此,每个直角三角形的面积为 A = b² / 2;无量纲常数为 1/2。现在将这个推理扩展到三维空间——找一个普通的金字塔(底边为正方形),将它与自身组合成一个简单的立体图形。

The easy case is easier to construct after we solve a similar but simpler problem: to find the area of a triangle with base b and height h. The area satisfies A ∼ hb, but what is the dimensionless constant? To find it, choose b and h to make an easy triangle: a right triangle with h = b. Two such triangles make an easy shape: a square with area b2. Thus each right triangle has area A = b2/2; the dimensionless constant is 1/2. Now extend this reasoning to three dimensions—find an ordinary pyramid (with a square base) that combines with itself to make an easy solid.

图像 什么是易固体?

What is the easy solid?

金字塔的方形底座暗示着一种方便的立体形状:也许每个底座都是立方体的一个面。这样,立方体就需要六个金字塔,它们的顶点在立方体的中心相交;因此,金字塔的长宽比为 h = b/2。为了简化计算,我们假设 b = 2 且 h = 1 满足此条件。

A convenient solid is suggested by the pyramid’s square base: Perhaps each base is one face of a cube. The cube then requires six pyramids whose tips meet in the center of the cube; thus the pyramids have the aspect ratio h = b/2. For numerical simplicity, let’s meet this condition with b = 2 and h = 1.

图像

六个这样的金字塔组成一个体积为b 3 = 8的立方体,因此一个金字塔的体积为4/3。由于每个金字塔的体积V ∼ hb 2,且这些金字塔的hb 2 = 4,因此V ∼ hb 2中的无量纲常数必定为1/3。因此,普通金字塔(a = 0的金字塔)的体积为V = hb 2 /3。

Six such pyramids form a cube with volume b3 = 8, so the volume of one pyramid is 4/3. Because each pyramid has volume V ∼ hb2, and hb2 = 4 for these pyramids, the dimensionless constant in V ∼ hb2 must be 1/3. The volume of an ordinary pyramid (a pyramid with a = 0) is therefore V = hb2/3.

问题 2.7 三角形底面

Problem 2.7 Triangular base

假设一个金字塔的高度为h,三角形底面面积为A,求其体积。假设金字塔的顶点位于底面质心正上方。然后尝试解答问题2.8

Guess the volume of a pyramid with height h and a triangular base of area A. Assume that the top vertex lies directly over the centroid of the base. Then try Problem 2.8.

问题 2.8 顶点位置

Problem 2.8 Vertex location

除非每个金字塔的顶顶点位于底面中心正上方,否则这六个金字塔无法构成一个立方体。因此,V = hb 2 /3 的结果可能仅适用于此限制。如果顶顶点位于某个底面顶点上方,那么它的体积是多少?

The six pyramids do not make a cube unless each pyramid’s top vertex lies directly above the center of the base. Thus the result V = hb2/3 might apply only with this restriction. If instead the top vertex lies above one of the base vertices, what is the volume?

前三个简单情况的测试预测 V = hb 2 /2(当 a = 0 时),而另一个简单情况 h = b/2(当 a = 0 时)则显示 V = hb 2 /3。这两种方法的预测结果相互矛盾。

The prediction from the first three easy-cases tests was V = hb2/2 (when a = 0), whereas the further easy case h = b/2 alongside a = 0 just showed that V = hb2/3. The two methods are making contradictory predictions.

图像 如何解决这一矛盾?

How can this contradiction be resolved?

矛盾肯定是在某个推理步骤中潜入的。要找到罪魁祸首,请依次回顾每个步骤。V ∝ h 的论证看起来正确。三个简单情况的条件——当a = 0 时V ∼ hb 2 ,当 b = 0 时 V ∼ ha 2,以及当 a = b 时 V = h(a 2 + b 2 )/2——看起来也正确。错误在于从这些约束条件直接跳到对任意 a 或 b 都有 V ∼ h(a 2 + b 2 ) 的预测。

The contradiction must have snuck in during one of the reasoning steps. To find the culprit, revisit each step in turn. The argument for V ∝ h looks correct. The three easy-case requirements—that V ∼ hb2 when a = 0, that V ∼ ha2 when b = 0, and that V = h(a2 + b2)/2 when a = b—also look correct. The mistake was leaping from these constraints to the prediction V ∼ h(a2 + b2) for any a or b.

相反,让我们尝试包含 ab 项的以下一般形式:

Instead let’s try the following general form that includes an ab term:

图像

然后通过重新应用简单情况的要求来求解系数 α、β 和 γ。

Then solve for the coefficients α, β, and γ by reapplying the easy-cases requirements.

b = 0 检验以及 h = b/2 的简单情况(证明对于普通金字塔,V = hb 2 /3)要求 α = 1/3。a = 0 检验同样要求 γ = 1/3。a = b 检验要求 α + β + γ = 1。因此 β = 1/3,瞧!

The b = 0 test along with the h = b/2 easy case, which showed that V = hb2/3 for an ordinary pyramid, require that α = 1/3. The a = 0 test similarly requires that γ = 1/3. And the a = b test requires that α + β + γ = 1. Therefore β = 1/3 and voilà,

图像

这个公式是比例推理、量纲分析和简单案例方法的结果,是精确的(问题 2.9)!

This formula, the result of proportional reasoning, dimensional analysis, and the method of easy cases, is exact (Problem 2.9)!

问题 2.9 积分

Problem 2.9 Integration

使用积分来证明 V = h(a 2 + ab + b 2 )/3。

Use integration to show that V = h(a2 + ab + b2)/3.

问题 2.10 截头三角锥

Problem 2.10 Truncated triangular pyramid

不要用正方形底座的金字塔,而是先用边长为b的等边三角形作为底座。然后用刀子平行于底座从顶部切下一块,做成截头立体。高度h顶边和底边长度分别为 a 和 b,这个立体的体积是多少?(另见问题 2.7。)

Instead of a pyramid with a square base, start with a pyramid with an equilateral triangle of side length b as its base. Then make the truncated solid by slicing a piece from the top using a knife parallel to the base. In terms of the height h and the top and bottom side lengths a and b, what is the volume of this solid? (See also Problem 2.7.)

问题 2.11 截头圆锥

Problem 2.11 Truncated cone

一个底面半径为 r 1、顶面半径为 r 2(顶面与底面平行)的圆台锥体的体积是多少?将你的公式推广到高为 h 的圆台锥体的体积,底面形状任意,面积为 A,相应的顶面面积为 A

What is the volume of a truncated cone with a circular base of radius r1 and circular top of radius r2 (with the top parallel to the base)? Generalize your formula to the volume of a truncated pyramid with height h, a base of an arbitrary shape and area Abase, and a corresponding top of area Atop.

2.4 流体力学:阻力

2.4 Fluid mechanics: Drag

前面的例子表明,简单情况可以检验并构造公式,但这些例子可以在没有简单情况的情况下完成(例如,使用微积分)。对于接下来的方程,由于流体力学中没有已知的精确解,因此简单情况和其他街头斗争工具几乎是取得进展的唯一途径。

The preceding examples showed that easy cases can check and construct formulas, but the examples can be done without easy cases (for example, with calculus). For the next equations, from fluid mechanics, no exact solutions are known in general, so easy cases and other street-fighting tools are almost the only way to make progress.

以下是流体力学的纳维-斯托克斯方程:

Here then are the Navier–Stokes equations of fluid mechanics:

图像

其中,v是流体的速度(作为位置和时间的函数),ρ 是其密度,p 是压力,ν 是运动粘度。这些方程描述了各种各样的现象,包括飞行、龙卷风和急流。

where v is the velocity of the fluid (as a function of position and time), ρ is its density, p is the pressure, and ν is the kinematic viscosity. These equations describe an amazing variety of phenomena including flight, tornadoes, and river rapids.

我们的例子是下面这个关于阻力的家庭实验。复印此页并将其放大2倍;然后剪下以下两个模板:

Our example is the following home experiment on drag. Photocopy this page while magnifying it by a factor of 2; then cut out the following two templates:

图像

用胶带将每个模板上的阴影部分粘在一起,做成一个圆锥体。两个圆锥体形状相同,但大圆锥体的高度和宽度是小圆锥体的两倍。

With each template, tape together the shaded areas to make a cone. The two resulting cones have the same shape, but the large cone has twice the height and width of the small cone.

图像

图像 当锥体朝下落下时,它们的终端速度(阻力平衡重量的速度)的近似比率是多少?

When the cones are dropped point downward, what is the approximate ratio of their terminal speeds (the speeds at which drag balances weight)?

纳维-斯托克斯方程包含了这个问题的答案。计算终端速度需要四个步骤。

The Navier–Stokes equations contain the answer to this question. Finding the terminal speed involves four steps.

步骤1. 施加边界条件。这些条件包括圆锥的运动以及不让流体进入纸张的要求。
第 2 步。 解该方程以及连续性方程图像= 0,以找到圆锥表面的压力和速度。
步骤3. 使用压力和速度来找到圆锥表面的压力和速度梯度;然后对产生的力进行积分以找到圆锥上的净力和扭矩。
步骤4. 利用净力和扭矩来计算圆锥的运动。这一步比较难,因为最终的运动必须与第一步假设的运动一致。如果不一致,就回到第一步,假设一个不同的运动,并祈祷到达这一步时运气更好。

遗憾的是,纳维-斯托克斯方程是耦合非线性偏微分方程。它们的解仅在非常简单的情况下才为人所知:例如,一个球体在粘性流体中以非常缓慢的速度运动,或者一个球体在零粘度流体中以任意速度运动。对于像柔性纸锥这样不规则、颤动形状的复杂流动,求解的希望微乎其微。

Unfortunately, the Navier–Stokes equations are coupled and nonlinear partial-differential equations. Their solutions are known only in very simple cases: for example, a sphere moving very slowly in a viscous fluid, or a sphere moving at any speed in a zero-viscosity fluid. There is little hope of solving for the complicated flow around an irregular, quivering shape such as a flexible paper cone.

问题 2.12 检查纳维-斯托克斯方程中的维度

Problem 2.12 Checking dimensions in the Navier–Stokes equations

检查纳维-斯托克斯方程的前三个项是否具有相同的维度。

Check that the first three terms of the Navier–Stokes equations have identical dimensions.

问题 2.13 运动粘度的维度

Problem 2.13 Dimensions of kinematic viscosity

根据纳维-斯托克斯方程,求出运动粘度 ν 的维数。

From the Navier–Stokes equations, find the dimensions of kinematic viscosity ν.

2.4.1 使用尺寸

2.4.1 Using dimensions

由于无法直接求解纳维-斯托克斯方程,我们不妨采用量纲分析法和一些简单案例。直接方法是利用这些方法推导出终端速度本身。间接方法是推导出阻力与下落速度的关系,然后找到阻力与锥体重量平衡的速度。这种两步法简化了问题。它只引入了一个新量(阻力),但省去了两个量:重力加速度和锥体的质量。

Because a direct solution of the Navier–Stokes equations is out of the question, let’s use the methods of dimensional analysis and easy cases. A direct approach is to use them to deduce the terminal velocity itself. An indirect approach is to deduce the drag force as a function of fall speed and then to find the speed at which the drag balances the weight of the cones. This two-step approach simplifies the problem. It introduces only one new quantity (the drag force) but eliminates two quantities: the gravitational acceleration and the mass of the cone.

问题 2.14 解释简化

Problem 2.14 Explaining the simplification

为什么阻力与重力加速度 g 和圆锥的质量 m 无关(但力却取决于圆锥的形状和大小)?

Why is the drag force independent of the gravitational acceleration g and of the cone’s mass m (yet the force depends on the cone’s shape and size)?

量纲原则是指有效方程中的所有项都具有相同的量纲。将其应用于阻力F,意味着在方程F = f(影响F的量)中,方程两边都具有力的量纲。因此,策略是先找到影响F的量,找到它们的量纲,然后将这些量合并成一个具有力的量纲的量。

The principle of dimensions is that all terms in a valid equation have identical dimensions. Applied to the drag force F, it means that in the equation F = f(quantities that affect F) both sides have dimensions of force. Therefore, the strategy is to find the quantities that affect F, find their dimensions, and then combine the quantities into a quantity with dimensions of force.

图像 阻力取决于哪些量,它们的量纲是多少?

On what quantities does the drag depend, and what are their dimensions?

图像

阻力取决于四个量:圆锥的两个参数和流体(空气)的两个参数。(关于ν的量纲,参见问题2.13。)

The drag force depends on four quantities: two parameters of the cone and two parameters of the fluid (air). (For the dimensions of ν, see Problem 2.13.)

图像 四个参数ν、r、ρν的任意组合是否都具有力的维度?

Do any combinations of the four parameters ν, r, ρ, and ν have dimensions of force?

下一步是将ν、r、ρ和ν组合成一个具有力的量纲的量。遗憾的是,可能性有很多——例如,

The next step is to combine ν, r, ρ, and ν into a quantity with dimensions of force. Unfortunately, the possibilities are numerous—for example,

图像

图像或者和的乘积组合图像。这些不好看的乘积之和也是一种力,因此阻力 F 可能是图像、,图像甚至更糟。

or the product combinations and . Any sum of these ugly products is also a force, so the drag force F could be , , or much worse.

缩小可能性需要一种比简单地猜测具有正确维度的组合更复杂的方法。为了开发这种复杂的方法,需要回到维度的第一原理:方程中的所有项都具有相同的维度。该原理适用于任何关于阻力的陈述,例如

Narrowing the possibilities requires a method more sophisticated than simply guessing combinations with correct dimensions. To develop the sophisticated approach, return to the first principle of dimensions: All terms in an equation have identical dimensions. This principle applies to any statement about drag such as

图像

其中,斑点 A、B 和 C 分别是 F、ν、r、ρ 和 ν 的函数。

where the blobs A, B, and C are functions of F, ν, r, ρ, and ν.

虽然这些斑点可能是极其复杂的函数,但它们的维度相同。因此,将每一项除以 A,得到公式

Although the blobs can be absurdly complex functions, they have identical dimensions. Therefore, dividing each term by A, which produces the equation

图像

使每个项都无量纲。同样的方法可以将任何有效方程转化为无量纲方程。因此,任何描述世界的(真实)方程都可以写成无量纲形式。

makes each term dimensionless. The same method turns any valid equation into a dimensionless equation. Thus, any (true) equation describing the world can be written in a dimensionless form.

任何无量纲形式都可以由无量纲群构建:即由变量的无量纲乘积构建。由于任何描述世界的方程都可以写成无量纲形式,而任何无量纲形式都可以用无量纲群来写出,因此任何描述世界的方程都可以用无量纲群来写出。

Any dimensionless form can be built from dimensionless groups: from dimensionless products of the variables. Because any equation describing the world can be written in a dimensionless form, and any dimensionless form can be written using dimensionless groups, any equation describing the world can be written using dimensionless groups.

图像 自由落体的例子(第 1.2 节)是否符合这一原则?

Is the free-fall example (Section 1.2) consistent with this principle?

在将此原理应用于复杂的阻力问题之前,先在自由落体这个简单的例子中尝试一下(第 1.2 节)。物体从高度 h 落下的精确撞击速度为 ν = ,图像其中 g 是重力加速度。这个结果实际上可以写成无量纲形式图像= 图像,而该形式本身只使用了无量纲组图像。新原理通过了初步检验。

Before applying this principle to the complicated problem of drag, try it in the simple example of free fall (Section 1.2). The exact impact speed of an object dropped from a height h is ν = where g is the gravitational acceleration. This result can indeed be written in the dimensionless form = , which itself uses only the dimensionless group . The new principle passes its first test.

这种公式的无量纲组分析,反过来就变成了一种综合方法。我们先来热身一下,综合一下冲击速度ν。首先,列出问题中的量;这里是ν、g和h。其次,将这些量组合成无量纲组。这里,所有无量纲组都可以从一个组构建出来。对于该组,我们选择ν2 / gh(具体选择不影响结论)。那么唯一可能的无量纲命题是

This dimensionless-group analysis of formulas, when reversed, becomes a method of synthesis. Let’s warm up by synthesizing the impact speed ν. First, list the quantities in the problem; here, they are ν, g, and h. Second, combine these quantities into dimensionless groups. Here, all dimensionless groups can be constructed just from one group. For that group, let’s choose ν2/gh (the particular choice does not affect the conclusion). Then the only possible dimensionless statement is

图像

(右侧是无量纲常数,因为那里没有第二个可用的组。)换句话说,ν 2 /gh ∼ 1 或 ν ∼ 图像

(The right side is a dimensionless constant because no second group is available to use there.) In other words, ν2/gh ∼ 1 or ν ∼ .

这一结果与1.2节中不太复杂的量纲分析的结果一致。事实上,如果只有一个无量纲群,两种分析方法都会得出相同的结论。然而,在诸如求阻力之类的难题中,不太复杂的方法无法以有用的形式提供其约束;这时,无量纲群方法就显得至关重要了。

This result reproduces the result of the less sophisticated dimensional analysis in Section 1.2. Indeed, with only one dimensionless group, either analysis leads to the same conclusion. However, in hard problems—for example, finding the drag force—the less sophisticated method does not provide its constraint in a useful form; then the method of dimensionless groups is essential.

问题 2.15 下降时间

Problem 2.15 Fall time

根据 g 和 h 综合出自由落体时间 t 的近似公式。

Synthesize an approximate formula for the free-fall time t from g and h.

问题 2.16 开普勒第三定律

Problem 2.16 Kepler’s third law

综合开普勒第三定律,将行星的轨道周期与其轨道半径联系起来。(另见问题 1.15。)

Synthesize Kepler’s third law connecting the orbital period of a planet to its orbital radius. (See also Problem 1.15.)

图像 对于阻力问题可以构建哪些无量纲群?

What dimensionless groups can be constructed for the drag problem?

一个无量纲群可以是 F/ρv 2 r 2;另一个群可以是 rν/ν 。任何其他群都可以由这些群构造(问题 2.17 ),因此该问题由两个独立的无量纲群描述。因此,最一般的无量纲表述是

One dimensionless group could be F/ρv2r2; a second group could be rν/ν. Any other group can be constructed from these groups (Problem 2.17), so the problem is described by two independent dimensionless groups. The most general dimensionless statement is then

图像

其中 f 是一个未知(但无量纲)的函数。

where f is a still-unknown (but dimensionless) function.

图像 哪个无量纲组属于左侧?

Which dimensionless group belongs on the left side?

目标是合成一个 F 的公式,而 F 仅出现在第一组 F/ρv 2 r 2中。考虑到这一限制,将第一组放在左侧,而不是将其包裹在仍然神秘的函数 f 中。基于此选择,关于阻力的最一般陈述是

The goal is to synthesize a formula for F, and F appears only in the first group F/ρv2r2. With that constraint in mind, place the first group on the left side rather than wrapping it in the still-mysterious function f. With this choice, the most general statement about drag force is

图像

圆锥体上(稳态)阻力的物理性质全部包含在无量纲函数 f 中。

The physics of the (steady-state) drag force on the cone is all contained in the dimensionless function f.

问题 2.17 只有两组

Problem 2.17 Only two groups

证明 F、ν、r、ρ 和图像仅产生两个独立的无量纲组。

Show that F, ν, r, ρ, and produce only two independent dimensionless groups.

问题2.18 一般有多少个组?

Problem 2.18 How many groups in general?

有没有一种通用的方法来预测独立无量纲群的数量?(答案在 1914 年由 Buckingham [ 9 ] 给出。)

Is there a general method to predict the number of independent dimensionless groups? (The answer was given in 1914 by Buckingham [9].)

这个过程看似毫无意义,因为它产生的阻力取决于未知函数 f。但它大大提高了我们找到 f 的概率。最初的问题公式需要猜测 F = h(ν, r, ρ, 图像) 中的四变量函数 h,而量纲分析将问题简化为猜测一个只有一个变量(νr/ 比值图像)的函数。统计学家兼物理学家 Harold Jeffreys 曾雄辩地描述过这种简化的价值(引自 [ 34 , 第82页]):

The procedure might seem pointless, having produced a drag force that depends on the unknown function f. But it has greatly improved our chances of finding f. The original problem formulation required guessing the four-variable function h in F = h(ν, r, ρ, ), whereas dimensional analysis reduced the problem to guessing a function of only one variable (the ratio νr/). The value of this simplification was eloquently described by the statistician and physicist Harold Jeffreys (quoted in [34, p. 82]):

一个好的单变量函数表可能需要一页纸;一个二变量函数表可能需要一卷书;一个三变量函数表可能需要一个书架;一个四变量函数表可能需要一个图书馆。

A good table of functions of one variable may require a page; that of a function of two variables a volume; that of a function of three variables a bookcase; and that of a function of four variables a library.

问题 2.19 截头金字塔的无量纲群

Problem 2.19 Dimensionless groups for the truncated pyramid

第 2.3 节中的截头金字塔的体积为

The truncated pyramid of Section 2.3 has volume

图像

用 V、h、a 和 b 组成无量纲群,并用这些群重写体积。(有很多方法可以做到这一点。)

Make dimensionless groups from V, h, a, and b, and rewrite the volume using these groups. (There are many ways to do so.)

2.4.2 使用简单案例

2.4.2 Using easy cases

尽管情况有所改善,但我们找到答案的机会看起来并不大:即使是单变量阻力问题也没有精确解。但在简单的情形下,它可能有精确解。由于最简单的情况往往是极端情况,所以先看看极端情况。

Although improved, our chances do not look high: Even the one-variable drag problem has no exact solution. But it might have exact solutions in its easy cases. Because the easiest cases are often extreme cases, look first at the extreme cases.

图像 极端情况有哪些?

Extreme cases of what?

未知函数 f 仅取决于 rν/ 图像

The unknown function f depends on only rν/,

图像

所以尝试 rν/ 的极端用法。图像然而,为了避免陷入盲目的符号推敲,首先要确定 rν/ 的含义图像。这个组合 rν/ 图像通常用 Re 表示,是著名的雷诺数。(其物理解释需要使用集中技术,详见第 3.4.3 节。)

so try extremes of rν/ However, to avoid lapsing into mindless symbol pushing, first determine the meaning of rν/. This combination rν/ , often denoted Re, is the famous Reynolds number. (Its physical interpretation requires the technique of lumping and is explained in Section 3.4.3.)

雷诺数通过未知函数 f 影响阻力:

The Reynolds number affects the drag force via the unknown function f:

图像

如果运气好的话,f 可以在雷诺数的极值处推导出来;如果运气更好的话,下落的锥体就是一个极端的例子。

With luck, f can be deduced at extremes of the Reynolds number; with further luck, the falling cones are an example of one extreme.

图像 下落的锥体是雷诺数的极值吗?

Are the falling cones an extreme of the Reynolds number?

雷诺数取决于r、ν和图像。对于速度ν,日常经验表明,锥体下落的速度约为1 ms −1(例如,在2倍以内)。尺寸r约为0.1 m(同样在2倍以内)。空气的运动粘度约为图像10 −5 m 2 s −1。雷诺数为

The Reynolds number depends on r, ν, and . For the speed ν, everyday experience suggests that the cones fall at roughly 1 ms−1 (within, say, a factor of 2). The size r is roughly 0.1 m (again within a factor of 2). And the kinematic viscosity of air is ∼ 10−5 m2 s−1. The Reynolds number is

图像

它明显大于1,因此下落锥体是高雷诺数的极端情况。(对于低雷诺数的情况,请尝试问题2.27并参见[ 38 ]。)

It is significantly greater than 1, so the falling cones are an extreme case of high Reynolds number. (For low Reynolds number, try Problem 2.27 and see [38].)

问题 2.20 日常流动中的雷诺数

Problem 2.20 Reynolds numbers in everyday flows

估算在水下巡航的潜艇、下落的花粉粒、下落的雨滴和横渡大西洋的 747 飞机的 Re。

Estimate Re for a submarine cruising underwater, a falling pollen grain, a falling raindrop, and a 747 crossing the Atlantic.

高雷诺数极限可以通过多种方式达到。一种方法是将粘度减小图像到0,因为图像位于雷诺数的分母中。因此,在高雷诺数极限下,粘度从问题中消失,阻力不应依赖于粘度。这种推理包含一些微妙的谬误,但其结论基本正确。(澄清这些微妙之处需要数学两个世纪的进步,最终导致了奇异摄动和边界层理论的出现[ 12,46 ]

The high-Reynolds-number limit can be reached many ways. One way is to shrink the viscosity to 0, because lives in the denominator of the Reynolds number. Therefore, in the limit of high Reynolds number, viscosity disappears from the problem and the drag force should not depend on viscosity. This reasoning contains several subtle untruths, yet its conclusion is mostly correct. (Clarifying the subtleties required two centuries of progress in mathematics, culminating in singular perturbations and the theory of boundary layers [12, 46].)

粘度仅通过雷诺数影响阻力:

Viscosity affects the drag force only through the Reynolds number:

图像

为了使 F 与粘度无关,F 必须与雷诺数无关!这样,问题就只包含一个独立的无量纲群,F/ρv 2 r 2,因此关于阻力的最一般表述是

To make F independent of viscosity, F must be independent of Reynolds number! The problem then contains only one independent dimensionless group, F/ρv2r2, so the most general statement about drag is

图像

阻力本身为 F ∼ ρv 2 r 2。由于 r 2与圆锥的横截面积 A 成正比,因此阻力通常写为

The drag force itself is then F ∼ ρv2r2. Because r2 is proportional to the cone’s cross-sectional area A, the drag force is commonly written

图像

虽然推导的是下落锥体,但只要雷诺数较高,结果也适用于任何物体。形状仅影响缺失的无量纲常数。对于球体,雷诺数大约为 1/4;对于垂直于其轴线运动的长圆柱体,雷诺数大约为 1/2;对于垂直于其表面运动的平板,雷诺数大约为 1。

Although the derivation was for falling cones, the result applies to any object as long as the Reynolds number is high. The shape affects only the missing dimensionless constant. For a sphere, it is roughly 1/4; for a long cylinder moving perpendicular to its axis, it is roughly 1/2; and for a flat plate moving perpendicular to its face, it is roughly 1.

2.4.3 终端速度

2.4.3 Terminal velocities

图像

得出 F ∼ ρv 2 A的结果足以预测圆锥体的终端速度。终端速度意味着加速度为零,因此阻力必须与重量平衡。重量为 W = σAg,其中 σ是纸的面密度(单位面积质量),A是模板切出四分之一截面后的面积。由于 A与横截面积 A 相当,因此重量粗略地定义为

The result F ∼ ρv2A is enough to predict the terminal velocities of the cones. Terminal velocity means zero acceleration, so the drag force must balance the weight. The weight is W = σpaperApaperg, where σpaper is the areal density of paper (mass per area) and Apaper is the area of the template after cutting out the quarter section. Because Apaper is comparable to the cross-sectional area A, the weight is roughly given by

图像

所以,

Therefore,

图像

面积除以终端速度

The area divides out and the terminal velocity becomes

图像

所有由同一张纸构成且形状相同的圆锥体,无论其大小如何,都以相同的速度下落!

All cones constructed from the same paper and having the same shape, whatever their size, fall at the same speed!

为了验证这个预测,我制作了第21页描述的大、小锥体,双手各举一个,高过头顶,然后让它们落下。它们从2米高处落下,持续了大约2秒,落地间隔不到0.1秒。低成本实验和低成本理论都证明了这一点!

To test this prediction, I constructed the small and large cones described on page 21, held one in each hand above my head, and let them fall. Their 2 m fall lasted roughly 2 s, and they landed within 0.1 s of one another. Cheap experiment and cheap theory agree!

问题 2.21 小锥体与大锥体的家庭实验

Problem 2.21 Home experiment of a small versus a large cone

亲自尝试一下锥形房屋实验(第 21 页)。

Try the cone home experiment yourself (page 21).

问题 2.22 四个堆叠圆锥体与一个圆锥体的家庭实验

Problem 2.22 Home experiment of four stacked cones versus one cone

预测比率

Predict the ratio

图像

测试你的预测。你能找到一种不需要计时设备的方法吗?

Test your prediction. Can you find a method not requiring timing equipment?

问题 2.23 估算终端速度

Problem 2.23 Estimating the terminal speed

估计或查找纸张的面密度;预测锥体的终端速度;然后将该预测与家庭实验的结果进行比较。

Estimate or look up the areal density of paper; predict the cones’ terminal speed; and then compare that prediction to the result of the home experiment.

2.5 总结和进一步的问题

2.5 Summary and further problems

正确的解决方案适用于所有情况,包括简单情况。因此,请在简单情况下检查任何提出的公式,并通过构建通过所有简单情况测试的表达式来猜测公式。要应用和扩展这些想法,请尝试以下问题,并参阅Cipra的简明易懂的书籍[ 10 ]。

A correct solution works in all cases, including the easy ones. Therefore, check any proposed formula in the easy cases, and guess formulas by constructing expressions that pass all easy-cases tests. To apply and extend these ideas, try the following problems and see the concise and instructive book by Cipra [10].

问题 2.24 栅栏柱错误

Problem 2.24 Fencepost errors

花园里有10米长的水平围栏,你想用竖立的立柱将其分成1米长的段。你需要10根还是11根竖立的立柱(包括两端的立柱)?

A garden has 10m of horizontal fencing that you would like to divide into 1m segments by using vertical posts. Do you need 10 or 11 vertical posts (including the posts needed at the ends)?

问题 2.25 奇数和

Problem 2.25 Odd sum

这是前 n 个奇数的和:

Here is the sum of the first n odd integers:

图像

a. 最后一项 l n等于 2n + 1 还是 2n − 1?

a. Does the last term ln equal 2n + 1 or 2n − 1?

b. 使用简单的情况来猜测 S n(作为 n 的函数)。

b. Use easy cases to guess Sn (as a function of n).

第 4.1 节讨论了另一种解决方案。

An alternative solution is discussed in Section 4.1.

问题 2.26 有初速度的自由落体

Problem 2.26 Free fall with initial velocity

1.2 节中的球静止释放。现在假设它被赋予初速度 ν 0(其中正 ν 0表示向上抛出)。猜想撞击速度 ν i

The ball in Section 1.2 was released from rest. Now imagine that it is given an initial velocity ν0 (where positive ν0 means an upward throw). Guess the impact velocity νi.

然后求解自由落体微分方程来找到精确的 ν i,并将精确结果与您的猜测进行比较。

Then solve the free-fall differential equation to find the exact νi, and compare the exact result to your guess.

问题 2.27 低雷诺数

Problem 2.27 Low Reynolds number

在极限 Re 图像1 中,猜测 f 的形式

In the limit Re 1, guess the form of f in

图像

该结果与正确的无量纲常数相结合,被称为斯托克斯阻力[ 12 ]。

The result, when combined with the correct dimensionless constant, is known as Stokes drag [12].

问题 2.28 范围公式

Problem 2.28 Range formula

一块石头在水平方向上(无空气阻力)能飞多远?利用尺寸和简单的情况,猜测射程 R 与发射速度 ν、发射角度 θ 和重力加速度 g 的关系公式。

How far does a rock travel horizontally (no air resistance)? Use dimensions and easy cases to guess a formula for the range R as a function of the launch velocity ν, the launch angle θ, and the gravitational acceleration g.

图像

问题 2.29 弹簧方程

Problem 2.29 Spring equation

理想质量-弹簧系统(第 3.4.2 节)的角频率为图像,其中 k 为弹簧常数,m 为质量。该表达式的分子中含有弹簧常数 k。使用 k 或 m 的极端情况来判断该位置是否正确。

The angular frequency of an ideal mass–spring system (Section 3.4.2) is , where k is the spring constant and m is the mass. This expression has the spring constant k in the numerator. Use extreme cases of k or m to decide whether that placement is correct.

问题 2.30 用胶带粘贴圆锥模板

Problem 2.30 Taping the cone templates

大圆锥模板(第21页)上的胶带标记宽度是小圆锥模板上胶带标记宽度的两倍。换句话说,如果大圆锥上的胶带宽度为6毫米,那么小圆锥上的胶带宽度应该是3毫米。为什么?

The tape mark on the large cone template (page 21) is twice as wide as the tape mark on the small cone template. In other words, if the tape on the large cone is, say, 6 mm wide, the tape on the small cone should be 3 mm wide. Why?

3

3

集结

Lumping

一颗绕行行星六个月后会在哪里?为了预测它的新位置,我们不能简单地将六个月乘以行星的当前速度,因为它的速度是不断变化的。微积分正是为此而发明的。它的基本思想是将时间划分成速度恒定的微小区间,将每个微小区间乘以相应的速度,计算出一个微小的距离,然后将这些微小的距离相加。

Where will an orbiting planet be 6 months from now? To predict its new location, we cannot simply multiply the 6 months by the planet’s current velocity, for its velocity constantly varies. Such calculations are the reason that calculus was invented. Its fundamental idea is to divide the time into tiny intervals over which the velocity is constant, to multiply each tiny time by the corresponding velocity to compute a tiny distance, and then to add the tiny distances.

令人惊讶的是,即使区间宽度无穷小,因此数量无限,这种计算通常也能精确完成。然而,符号运算可能非常冗长,更糟糕的是,即使问题的微小变化也常常导致计算无法进行。例如,使用微积分方法,我们可以精确计算高斯函数 e −x 2在 x = 0 和 ∞ 之间的面积;然而,如果任一极限为零或无穷大以外的任何值,精确计算就变得不可能了。

Amazingly, this computation can often be done exactly, even when the intervals have infinitesimal width and are therefore infinite in number. However, the symbolic manipulations can be lengthy and, worse, are often rendered impossible by even small changes to the problem. Using calculus methods, for example, we can exactly calculate the area under the Gaussian e−x2 between x = 0 and ∞; yet if either limit is any value except zero or infinity, an exact calculation becomes impossible.

相比之下,近似方法稳健:它们几乎总能给出合理的答案。准确度最低但稳健性最高的方法是集中法。与其将变化过程分解成许多小部分,不如将其组合或集中成一两个部分。我们将通过从人口统计学(第 3.1 节)到非线性微分方程(第 3.5 节)等示例来说明这种简单的近似方法及其优势。

In contrast, approximate methods are robust: They almost always provide a reasonable answer. And the least accurate but most robust method is lumping. Instead of dividing a changing process into many tiny pieces, group or lump it into one or two pieces. This simple approximation and its advantages are illustrated using examples ranging from demographics (Section 3.1) to nonlinear differential equations (Section 3.5).

3.1 估计人口:有多少婴儿?

3.1 Estimating populations: How many babies?

第一个例子是估算美国的婴儿数量。为了更明确起见,我们将婴儿(直到他或她2岁)称为婴儿。精确计算需要每个美国人的出生日期。美国人口普查局每十年收集一次此类信息或类似的信息。

The first example is to estimate the number of babies in the United States. For definiteness, call a child a baby until he or she turns 2 years old. An exact calculation requires the birth dates of every person in the United States. This, or closely similar, information is collected once every decade by the US Census Bureau.

为了近似这些海量数据,美国人口普查局[ 47 ]公布了各年龄段的人口数量。1991年的数据是一组位于一条曲线N(t)上的点,其中t代表年龄。那么

As an approximation to this voluminous data, the Census Bureau [47] publishes the number of people at each age. The data for 1991 is a set of points lying on a wiggly line N(t), where t is age. Then

图像

图片

问题 3.1 垂直轴的尺寸

Problem 3.1 Dimensions of the vertical axis

为什么纵轴的单位是“人/年”,而不是“人”?同样地,为什么纵轴的维度是 T −1

Why is the vertical axis labeled in units of people per year rather than in units of people? Equivalently, why does the axis have dimensions of T−1?

这种方法存在几个问题。首先,它依赖于美国人口普查局的庞大资源,因此无法在荒岛上进行粗略计算。其次,它需要对一条没有解析形式的曲线进行积分,因此积分必须通过数值计算完成。第三,积分是针对特定问题的数据,而数学应该关注的是普遍性。简而言之,精确积分几乎无法提供任何洞见,转移价值也微乎其微。与其精确积分人口曲线,不如近似积分——将曲线集中到一个矩形中。

This method has several problems. First, it depends on the huge resources of the US Census Bureau, so it is not usable on a desert island for back-of-the-envelope calculations. Second, it requires integrating a curve with no analytic form, so the integration must be done numerically. Third, the integral is of data specific to this problem, whereas mathematics should be about generality. An exact integration, in short, provides little insight and has minimal transfer value. Instead of integrating the population curve exactly, approximate it—lump the curve into one rectangle.

图像 这个矩形的高度和宽度是多少?

What are the height and width of this rectangle?

矩形的宽度代表一个时间,而与人口相关的一个合理时间是预期寿命。它大约是80年,所以假设每个人都在80岁生日那天突然去世,这样宽度就等于80年。矩形的高度可以根据矩形的面积计算出来,面积就是美国的人口——2008年恰好是3亿。因此,

The rectangle’s width is a time, and a plausible time related to populations is the life expectancy. It is roughly 80 years, so make 80 years the width by pretending that everyone dies abruptly on his or her 80th birthday. The rectangle’s height can be computed from the rectangle’s area, which is the US population—conveniently 300 million in 2008. Therefore,

图片

图像 为什么预期寿命从80岁下降75岁?

Why did the life expectancy drop from 80 to 75 years?

捏造预期寿命简化了心算除法:75 很容易被 3 和 300 整除。其误差并不大于合并造成的误差,甚至可能抵消合并误差。以 75 年为宽度,高度约为 4 × 10 6−1

Fudging the life expectancy simplifies the mental division: 75 divides easily into 3 and 300. The inaccuracy is no larger than the error made by lumping, and it might even cancel the lumping error. Using 75 years as the width makes the height approximately 4 × 106 yr−1.

图片

对 t = 0 . . . 2 年范围内的人口曲线进行积分就变成了乘法:

Integrating the population curve over the range t = 0 . . . 2 yr becomes just multiplication:

图片

人口普查局的数据非常接近:7.980 × 10 6。合并造成的误差抵消了将预期寿命捏造为 75 岁造成的误差!

The Census Bureau’s figure is very close: 7.980 × 106. The error from lumping canceled the error from fudging the life expectancy to 75 years!

问题3.2 垃圾填埋场体积

Problem 3.2 Landfill volume

估算美国每年填埋的一次性尿布的体积。

Estimate the US landfill volume used annually by disposable diapers.

问题 3.3 行业收入

Problem 3.3 Industry revenues

估计美国尿布行业的年收入。

Estimate the annual revenue of the US diaper industry.

3.2 估算积分

3.2 Estimating integrals

美国人口曲线(第 3.1 节)难以积分,部分原因是它是未知的。但即使是众所周知的函数也可能难以积分。在这种情况下,两种集中方法特别有用:1/e 启发式(第 3.2.1 节)和半峰全宽 (FWHM) 启发式(第 3.2.2 节)。

The US population curve (Section 3.1) was difficult to integrate partly because it was unknown. But even well-known functions can be difficult to integrate. In such cases, two lumping methods are particularly useful: the 1/e heuristic (Section 3.2.1) and the full width at half maximum (FWHM) heuristic (Section 3.2.2).

3.2.1 1/e启发式

3.2.1 1/e heuristic

电子电路、大气压力和放射性衰变包含普遍存在的指数及其积分(此处以无量纲形式给出)

Electronic circuits, atmospheric pressure, and radioactive decay contain the ubiquitous exponential and its integral (given here in dimensionless form)

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为了近似它的值,我们将 e −t曲线集中到一个矩形中。

To approximate its value, let’s lump the e−t curve into one rectangle.

图像 矩形的宽度和高度应该选择什么值?

What values should be chosen for the width and height of the rectangle?

矩形的合理高度是 e −t的最大值,即 1。要选择其宽度,请使用显著变化作为标准(第 3.3.3 节中再次使用的方法):选择 e −t的显著变化;然后找到产生此变化的宽度 Δt。在指数衰减中,一个简单而自然的显著变化是当 e −t成为 e 的一个因子,更接近其最终值(这里为 0,因为 t 趋于 ∞)。根据此标准,Δt = 1。则集中矩形具有单位面积——它是积分的精确值!

A reasonable height for the rectangle is the maximum of e−t, namely 1. To choose its width, use significant change as the criterion (a method used again in Section 3.3.3): Choose a significant change in e−t; then find the width Δt that produces this change. In an exponential decay, a simple and natural significant change is when e−t becomes a factor of e closer to its final value (which is 0 here because t goes to ∞). With this criterion, Δt = 1. The lumping rectangle then has unit area—which is the exact value of the integral!

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受到这个结果的鼓舞,让我们在困难的积分上尝试启发式方法

Encouraged by this result, let’s try the heuristic on the difficult integral

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再次将面积集中到一个矩形中。它的高度是 e −x ²的最大值,即 1。它的宽度足以使 e −x ²下降 e 倍。这个下降发生在 x = ±1 处,因此宽度为 Δx = 2,面积为 1 × 2。精确面积约为图片1.77(见 2.1 节),因此集中计算的误差仅为 13%:对于如此简短的推导过程,精度极高。

Again lump the area into a single rectangle. Its height is the maximum of e−x2 , which is 1. Its width is enough that e−x2 falls by a factor of e. This drop happens at x = ±1, so the width is Δx = 2 and its area is 1 × 2. The exact area is ≈ 1.77 (Section 2.1), so lumping makes an error of only 13%: For such a short derivation, the accuracy is extremely high.

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问题 3.4 一般指数衰减

Problem 3.4 General exponential decay

使用集中法来估计积分

Use lumping to estimate the integral

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使用维度分析和简单案例来检查你的答案是否合理。

Use dimensional analysis and easy cases to check that your answer makes sense.

问题 3.5 大气压力

Problem 3.5 Atmospheric pressure

大气密度ρ随高度z大致呈指数衰减:

Atmospheric density ρ decays roughly exponentially with height z:

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其中 ρ 0是海平面密度,H 是所谓的标高(密度下降 e 倍的高度)。您可以利用日常经验估算 H。

where ρ0 is the density at sea level, and H is the so-called scale height (the height at which the density falls by a factor of e). Use your everyday experience to estimate H.

然后通过估算无限高的空气圆柱体的重量来估算海平面的大气压力。

Then estimate the atmospheric pressure at sea level by estimating the weight of an infinitely high cylinder of air.

问题 3.6 圆锥自由落体距离

Problem 3.6 Cone free-fall distance

2.4 节中的圆锥体在达到其终端速度的相当一部分之前,大概会下落多远?与 2 米的下落高度相比,这个距离是多少?提示:粗略地画出圆锥体的加速度与时间的关系图,并进行集中近似。

Roughly how far does a cone of Section 2.4 fall before reaching a significant fraction of its terminal velocity? How large is that distance compared to the drop height of 2 m? Hint: Sketch (very roughly) the cone’s acceleration versus time and make a lumping approximation.

3.2.2 半峰全宽

3.2.2 Full width at half maximum

另一种合理的集中启发法出现在光谱学的早期。当光谱仪扫描一系列波长时,图表记录仪会绘制出分子吸收该波长辐射的强度。这条曲线包含许多峰,它们的位置和面积揭示了分子的结构(这对发展量子理论至关重要[ 14 ])。但在数字图表记录仪出现的几十年前,这些峰的面积是如何计算的呢?

Another reasonable lumping heuristic arose in the early days of spectroscopy. As a spectroscope swept through a range of wavelengths, a chart recorder would plot how strongly a molecule absorbed radiation of that wavelength. This curve contains many peaks whose location and area reveal the structure of the molecule (and were essential in developing quantum theory [14]). But decades before digital chart recorders existed, how could the areas of the peaks be computed?

它们是通过将峰集中到一个矩形中来计算的,矩形的高度为峰高,宽度为半峰全宽 (FWHM)。1/e 启发式方法使用因子 e 作为显著变化,而 FWHM 启发式方法使用因子 2。

They were computed by lumping the peak into a rectangle whose height is the height of the peak and whose width is the full width at half maximum (FWHM). Where the 1/e heuristic uses a factor of e as the significant change, the FWHM heuristic uses a factor of 2.

在高斯积分上尝试这个方法图片

Try this recipe on the Gaussian integral .

e −x 2的最大高度为 1,因此半峰位于 x = ± 处图片,全宽为 2 图片。因此,集总矩形的面积 2 图片≈ 1.665。精确面积约为图片1.77(见第 2.1 节):FWHM 启发式方法的误差仅为 6%,大约是 1/e 启发式方法误差的一半。

The maximum height of e−x2 is 1, so the half maxima are at x = ± and the full width is 2. The lumped rectangle therefore has area 2 ≈ 1.665. The exact area is ≈ 1.77 (Section 2.1): The FWHM heuristic makes an error of only 6%, which is roughly one-half the error of the 1/e heuristic.

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问题 3.7 尝试 FWHM 启发式

Problem 3.7 Trying the FWHM heuristic

对下列积分进行单矩形集中估计。使用 FWHM 启发式方法选择矩形的高度和宽度。每个估计值的准确度如何?

Make single-rectangle lumping estimates of the following integrals. Choose the height and width of the rectangle using the FWHM heuristic. How accurate is each estimate?

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3.2.3 斯特林近似

3.2.3 Stirling’s approximation

接下来,1/e 和 FWHM 集中启发式方法帮助我们近似普遍存在的阶乘函数 n!;该函数的用途广泛,从概率论到统计力学,再到算法分析。对于正整数,n! 定义为 n × (n − 1) × (n − 2) × · · · × 2 × 1。在这种离散形式下,它很难近似。然而,n! 的积分表示,

The 1/e and FWHM lumping heuristics next help us approximate the ubiquitous factorial function n!; this function’s uses range from probability theory to statistical mechanics and the analysis of algorithms. For positive integers, n! is defined as n × (n − 1) × (n − 2) × · · · × 2 × 1. In this discrete form, it is difficult to approximate. However, the integral representation for n!,

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即使 n 不是正整数,也能提供定义——并且可以使用集中法来近似该积分。

provides a definition even when n is not a positive integer—and this integral can be approximated using lumping.

集中分析将生成几乎所有斯特林著名的近似公式

The lumping analysis will generate almost all of Stirling’s famous approximation formula

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图像 集中需要一个峰值,但被积函数t n e −t 有峰值吗?

Lumping requires a peak, but does the integrand tne−t have a peak?

要理解被积函数 t n e −t或 t n /e t,需要考察 t 的极端情况。当 t = 0 时,被积函数为 0。在另一个极端,即 t → ∞ 时,多项式因子 t n使乘积无穷大,而指数因子 e −t使其为零。谁会胜出呢?e t的泰勒级数包含 t 的所有幂(且系数为正),因此它是一个递增的无限次多项式。因此,当 t 趋于无穷大时,e t会超越任何多项式 t n,并使被积函数 t n /e t在 t → ∞ 的极端情况下等于 0。由于被积函数在两个极端都为零,因此在这两个极端之间必然有一个峰值。事实上,它只有一个峰值。(你能证明这一点吗?)

To understand the integrand tne−t or tn/et, examine the extreme cases of t. When t = 0, the integrand is 0. In the opposite extreme, t → ∞, the polynomial factor tn makes the product infinity while the exponential factor e−t makes it zero. Who wins that struggle? The Taylor series for et contains every power of t (and with positive coefficients), so it is an increasing, infinite-degree polynomial. Therefore, as t goes to infinity, et outruns any polynomial tn and makes the integrand tn/et equal 0 in the t → ∞ extreme. Being zero at both extremes, the integrand must have a peak in between. In fact, it has exactly one peak. (Can you show that?)

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随着 n 的增加,多项式因子 t n 的强度会增强,因此 t n会一直存在,直到 t 更高时,e t才会超过它。因此,t n /e t的峰值会随着 n 的增加而右移。该图证实了这一预测,并表明峰值出现在 t = n 时。让我们用微积分来验证一下,最大化 t n e −t,或者更简单地说,最大化其对数 f(t) = n ln t − t。在峰值处,函数的斜率为零。由于 df/dt = n/t−1,峰值出现在 t peak = n 时,此时被积函数 t n e −t为 n n e −n——从而重现了斯特林公式中最大且最重要的因子。

Increasing n strengthens the polynomial factor tn, so tn survives until higher t before et outruns it. Therefore, the peak of tn/et shifts right as n increases. The graph confirms this prediction and suggests that the peak occurs at t = n. Let’s check by using calculus to maximize tne−t or, more simply, to maximize its logarithm f(t) = n ln t − t. At a peak, a function has zero slope. Because df/dt = n/t−1, the peak occurs at tpeak = n, when the integrand tne−t is nne−n—thus reproducing the largest and most important factor in Stirling’s formula.

图像 什么是合理的集中矩形?

What is a reasonable lumping rectangle?

矩形的高度是峰值高度 n n e −n。对于矩形的宽度,可以使用 1/e 或 FWHM 启发式方法。由于这两种启发式方法都需要近似 t n e −t,因此将其对数 f(t) 围绕 t = n 时的峰值以泰勒级数展开:

The rectangle’s height is the peak height nne−n. For the rectangle’s width, use either the 1/e or the FWHM heuristics. Because both heuristic require approximating tne−t, expand its logarithm f(t) in a Taylor series around its peak at t = n:

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泰勒展开式的第二项为零,因为 f(t) 在峰值处的斜率为零。第三项中,t = n 时的二阶导数 d ² f/dt ²为 −n/t ²或 −1/n。因此,

The second term of the Taylor expansion vanishes because f(t) has zero slope at the peak. In the third term, the second derivative d2f/dt2 at t = n is −n/t2 or −1/n. Thus,

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要将 t n e −t减小F 倍,需要将 f(t) 减小 ln F。这意味着图片。由于矩形的宽度为 2Δt,因此 n! 的集总面积估计值为

To decrease tne−t by a factor of F requires decreasing f(t) by ln F. This choice means . Because the rectangle’s width is 2Δt, the lumped-area estimate of n! is

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相比之下,斯特林公式为 n! ≈ n n e −n 图片。Lumping 几乎解释了所有因子。n n e −n因子是矩形的高度,而图片因子是矩形的宽度。虽然确切的图片因子仍然未知(问题 3.9),但它的近似值在 13%(1/e 启发式)或 6%(FWHM 启发式)以内。

For comparison, Stirling’s formula is n! ≈ nne−n Lumping has explained almost every factor. The nne−n factor is the height of the rectangle, and the factor is from the width of the rectangle. Although the exact factor remains mysterious (Problem 3.9), it is approximated to within 13% (the 1/e heuristic) or 6% (the FWHM heuristic).

问题 3.8 巧合吗?

Problem 3.8 Coincidence?

高斯分布下面积的 FWHM 近似值(第 3.2.2 节)也精确到 6%。巧合吗?

The FWHM approximation for the area under a Gaussian (Section 3.2.2) was also accurate to 6%. Coincidence?

问题 3.9 斯特林公式中的精确常数

Problem 3.9 Exact constant in Stirling’s formula

更精确的常数因子从何图片而来?

Where does the more accurate constant factor of come from?

3.3 估计导数

3.3 Estimating derivatives

在前面的例子中,集中法有助于估计积分。由于积分和微分密切相关,集中法也提供了一种估算导数的方法。该方法始于对导数进行量纲观察。导数是微分的比值;例如,df/dx 是 df 与 dx 的比值。由于 d 是无量纲的(见第 1.3.2 节),因此 df/dx 的量纲就是 f/x 的量纲。这个有用而又令人惊奇的结论值得用一个熟悉的例子来检验:对高度 y 求时间 t 的微分,得到速度 dy/dt,其 LT −1的量纲实际上就是 y/t 的量纲。

In the preceding examples, lumping helped estimate integrals. Because integration and differentiation are closely related, lumping also provides a method for estimating derivatives. The method begins with a dimensional observation about derivatives. A derivative is a ratio of differentials; for example, df/dx is the ratio of df to dx. Because d is dimensionless (Section 1.3.2), the dimensions of df/dx are the dimensions of f/x. This useful, surprising conclusion is worth testing with a familiar example: Differentiating height y with respect to time t produces velocity dy/dt, whose dimensions of LT−1 are indeed the dimensions of y/t.

问题 3.10 二阶导数的维数

Problem 3.10 Dimensions of a second derivative

d 2 f/dx 2的尺寸是多少?

What are the dimensions of d2f/dx2?

3.3.1 正割近似

3.3.1 Secant approximation

由于 df/dx 和 f/x 具有相同的量纲,因此它们的大小可能相似:

As df/dx and f/x have identical dimensions, perhaps their magnitudes are similar:

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从几何学上讲,导数 df/dx 是切线的斜率,而近似值 f/x 是割线的斜率。通过用割线代替曲线,我们得到了一个集中近似。

Geometrically, the derivative df/dx is the slope of the tangent line, whereas the approximation f/x is the slope of the secant line. By replacing the curve with the secant line, we make a lumping approximation.

让我们用一个简单的函数(例如 f(x) = x 2 )来测试这个近似值。好消息是,正割和正切的斜率仅相差 2 倍:

Let’s test the approximation on an easy function such as f(x) = x2. Good news—the secant and tangent slopes differ only by a factor of 2:

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问题 3.11 高阶幂

Problem 3.11 Higher powers

研究 f(x) = x n的正割近似值。

Investigate the secant approximation for f(x) = xn.

问题 3.12 二阶导数

Problem 3.12 Second derivatives

使用正割近似值来估计 d 2 f/dx 2(其中 f(x) = x 2 )。该近似值与精确二阶导数相比如何?

Use the secant approximation to estimate d2f/dx2 with f(x) = x2. How does the approximation compare to the exact second derivative?

图像 f(x) = x 2 + 100的正割近似值有多准确

How accurate is the secant approximation for f(x) = x2 + 100?

正割近似快速且实用,但可能会产生较大的误差。例如,当 f(x) = x 2 + 100 时,x = 1 处的正割和正切斜率差异巨大。正切斜率 df/dx 为 2,而正割斜率 f(1)/1 为 101。这两个斜率的比值虽然无量纲,但却大得令人担忧。

The secant approximation is quick and useful but can make large errors. When f(x) = x2 + 100, for example, the secant and tangent at x = 1 have dramatically different slopes. The tangent slope df/dx is 2, whereas the secant slope f(1)/1 is 101. The ratio of these two slopes, although dimensionless, is distressingly large.

问题 3.13 调查差异

Problem 3.13 Investigating the discrepancy

以 f(x) = x 2 + 100 为公式,画出

With f(x) = x2 + 100, sketch the ratio

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作为 x 的函数。这个比率不是常数!为什么无量纲因子不是常数?(这个问题很棘手。)

as a function of x. The ratio is not constant! Why is the dimensionless factor not constant? (That question is tricky.)

替换导数 df/dx 的巨大差异在于

The large discrepancy in replacing the derivative df/dx, which is

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割线斜率 f(x)/x 的近似值源于两个近似值。第一个近似值取 Δx = x 而不是 Δx = 0。则 df/dx ≈ (f(x) − f(0))/x。第一个近似值得出从 (0, f(0)) 到 (x, f(x)) 的直线的斜率。第二个近似值将 f(0) 替换为 0,得出 df/dx ≈ f/x;该比率是从 (0, 0) 到 (x, f(x)) 的割线的斜率。

with the secant slope f(x)/x is due to two approximations. The first approximation is to take Δx = x rather than Δx = 0. Then df/dx ≈ (f(x) − f(0))/x. This first approximation produces the slope of the line from (0, f(0)) to (x, f(x)). The second approximation replaces f(0) with 0, which produces df/dx ≈ f/x; that ratio is the slope of the secant from (0, 0) to (x, f(x)).

3.3.2 改进的割线近似

3.3.2 Improved secant approximation

第二个近似值是通过从 (0, f(0)) 而不是 (0, 0) 开始割线来解决的。

The second approximation is fixed by starting the secant at (0, f(0)) instead of (0, 0).

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图像 经过这种变化,当f(x) = x 2 + C时,正割和正切的斜率是多少

With that change, what are the secant and tangent slopes when f(x) = x2 + C?

将从 (0, 0) 开始的割线称为原点割线;将新的割线称为 x = 0 割线。这样,无论常数 C 是多少,x = 0 割线的斜率始终是正切的一半。x = 0 割线近似值对垂直平移具有鲁棒性,即不受其影响。

Call the secant starting at (0, 0) the origin secant; call the new secant the x = 0 secant. Then the x = 0 secant always has one-half the slope of the tangent, no matter the constant C. The x = 0 secant approximation is robust against—is unaffected by—vertical translation.

图像 x = 0正割近似对于水平平移的稳健性如何

How robust is the x = 0 secant approximation against horizontal translation?

为了研究 x = 0 正割如何处理水平平移,将 f(x) = x 2向右平移 100 度,得到 f(x) = (x−100) 2。在抛物线的顶点 x = 100,x = 0 的割线,从 (0, 10 4 ) 到 (100, 0),斜率为 -100;然而,切线的斜率为零。因此,x = 0 的割线虽然比原点割线有所改进,但仍然会受到水平平移的影响。

To investigate how the x = 0 secant handles horizontal translation, translate f(x) = x2 rightward by 100 to make f(x) = (x−100)2. At the parabola’s vertex x = 100, the x = 0 secant, from (0, 104) to (100, 0), has slope −100; however, the tangent has zero slope. Thus the x = 0 secant, although an improvement on the origin secant, is affected by horizontal translation.

3.3.3 显著变化近似值

3.3.3 Significant-change approximation

导数本身不受水平和垂直平移的影响,因此适当近似的导数可能具有平移不变性。近似导数为

The derivative itself is unaffected by horizontal and vertical translation, so a derivative suitably approximated might be translation invariant. An approximate derivative is

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其中 Δx 不为零,但仍然很小。

where Δx is not zero but is still small.

图像 Δx应该有多小 Δx = 0.01够小吗?

How small should Δx be? Is Δx = 0.01 small enough?

选择 Δx = 0.01 有两个缺陷。首先,当 x 有维度时,它不起作用。如果 x 是长度,那么多长才足够小?选择 Δx = 1 毫米对于计算与太阳系相关的导数来说可能足够小,但对于计算与下落雾滴相关的导数来说可能太大。其次,没有固定的选择能够保持尺度不变。虽然 Δx = 0.01 在 f(x) = sin x 时可以产生精确的导数,但当 f(x) = sin 1000x(即简单地将 x 缩放到 1000x 的结果)时,它就失效了。

The choice Δx = 0.01 has two defects. First, it cannot work when x has dimensions. If x is a length, what length is small enough? Choosing Δx = 1 mm is probably small enough for computing derivatives related to the solar system, but is probably too large for computing derivatives related to falling fog droplets. Second, no fixed choice can be scale invariant. Although Δx = 0.01 produces accurate derivatives when f(x) = sin x, it fails when f(x) = sin 1000x, the result of simply rescaling x to 1000x.

这些问题建议尝试以下显著变化近似:

These problems suggest trying the following significant-change approximation:

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因为这里的 Δx 由感兴趣点处的曲线属性定义,而不偏向特定的坐标值或 Δx 值,所以近似值是尺度和平移不变的。

Because the Δx here is defined by the properties of the curve at the point of interest, without favoring particular coordinate values or values of Δx, the approximation is scale and translation invariant.

为了说明这个近似值,我们尝试 f(x) = cos x,并用三个近似值估计 x = 3π/2 处的 df/dx:原点割线、x = 0 割线和显著变化近似。原点割线从 (0, 0) 到 (3π/2, 0),因此斜率为零。它与精确斜率 1 的近似值很差。x = 0正割从 (0, 1) 到 (3π/2, 0),因此它的斜率为 −2/3π,这比预测零斜率更糟糕,因为甚至符号都是错误的!

To illustrate this approximation, let’s try f(x) = cos x and estimate df/dx at x = 3π/2 with the three approximations: the origin secant, the x = 0 secant, and the significant-change approximation. The origin secant goes from (0, 0) to (3π/2, 0), so it has zero slope. It is a poor approximation to the exact slope of 1. The x = 0 secant goes from (0, 1) to (3π/2, 0), so it has a slope of −2/3π, which is worse than predicting zero slope because even the sign is wrong!

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显著变化近似值可能更准确。f(x) = cos x 的显著变化是多少?由于余弦值变化了 2(从 −1 到 1),所以称 1/2 为 f(x) 的显著变化。当 x 从 3π/2(其中 f(x) = 0)变为 3π/2 + π/6(其中 f(x) = 1/2)时,就会发生这种变化。换句话说,Δx 等于 π/6。因此,近似导数为

The significant-change approximation might provide more accuracy. What is a significant change in f(x) = cos x? Because the cosine changes by 2 (from −1 to 1), call 1/2 a significant change in f(x). That change happens when x changes from 3π/2, where f(x) = 0, to 3π/2 + π/6, where f(x) = 1/2. In other words, Δx is π/6. The approximate derivative is therefore

图像

该估计值约为 0.955 — 非常接近 1 的真实导数。

This estimate is approximately 0.955—amazingly close to the true derivative of 1.

问题 3.14 二次函数的导数

Problem 3.14 Derivative of a quadratic

设 f(x) = x 2,使用三个近似值估计 x = 5 处的 df/dx:原点正割、x = 0 正割和显著变化近似值。将这些估计值与真实斜率进行比较。

With f(x) = x2, estimate df/dx at x = 5 using three approximations: the origin secant, the x = 0 secant, and the significant-change approximation. Compare these estimates to the true slope.

问题 3.15 对数的导数

Problem 3.15 Derivative of the logarithm

使用显著变化近似值来估计 x = 10 处的 ln x 的导数。将估计值与真实斜率进行比较。

Use the significant-change approximation to estimate the derivative of ln x at x = 10. Compare the estimate to the true slope.

问题 3.16 Lennard–Jones 势

Problem 3.16 Lennard–Jones potential

Lennard-Jones势是两个非极性分子(例如N 2或CH 4 )之间相互作用能的模型。其形式为

The Lennard–Jones potential is a model of the interaction energy between two nonpolar molecules such as N2 or CH4. It has the form

图像

其中 r 是分子间的距离,є 和 σ 是依赖于分子的常数。利用原点割线估计 r 0 ,即 V(r) 最小时的分离 r 。将估计值与使用微积分计算得出的真实 r 0进行比较。

where r is the distance between the molecules, and є and σ are constants that depend on the molecules. Use the origin secant to estimate r0, the separation r at which V(r) is a minimum. Compare the estimate to the true r0 found using calculus.

问题 3.17 近似最大值和最小值

Problem 3.17 Approximate maxima and minima

设 f(x) 为增函数,g(x) 为减函数。利用原点正割近似证明 h(x) = f(x) + g(x) 存在最小值,此时 f(x) = g(x)。这条经验法则推广了问题 3.16,通常被称为平衡启发式。

Let f(x) be an increasing function and g(x) a decreasing function. Use the origin secant to show, approximately, that h(x) = f(x) + g(x) has a minimum where f(x) = g(x). This useful rule of thumb, which generalizes Problem 3.16, is often called the balancing heuristic.

3.4 分析微分方程:弹簧质量系统

3.4 Analyzing differential equations: The spring–mass system

估计导数将微分简化为除法(第 3.3 节);从而将微分方程简化为代数方程。

Estimating derivatives reduces differentiation to division (Section 3.3); it thereby reduces differential equations to algebraic equations.

图像

为了生成一个示例方程进行分析,将一个质量为 m 的块连接到一个弹簧常数(刚度)为 k 的理想弹簧上,将该块相对于平衡位置 x = 0 向右拉动 x 0的距离,并在时间 t = 0 时释放它。块来回振荡,其位置 x 由理想弹簧微分方程描述

To produce an example equation to analyze, connect a block of mass m to an ideal spring with spring constant (stiffness) k, pull the block a distance x0 to the right relative to the equilibrium position x = 0, and release it at time t = 0. The block oscillates back and forth, its position x described by the ideal-spring differential equation

图像

让我们近似该方程并由此估计振荡频率。

Let’s approximate the equation and thereby estimate the oscillation frequency.

3.4.1 检查尺寸

3.4.1 Checking dimensions

看到任何方程式时,首先要检查它的量纲(第一章)。如果所有项的量纲不一致,则该方程式不值得求解——这将大大节省求解的精力。如果量纲一致,则检查结果会促使你思考这些项的含义;这种思考有助于为求解方程式和理解任何解做好准备。

Upon seeing any equation, first check its dimensions (Chapter 1). If all terms do not have identical dimensions, the equation is not worth solving—a great savings of effort. If the dimensions match, the check has prompted reflection on the meaning of the terms; this reflection helps prepare for solving the equation and for understanding any solution.

图像 弹簧方程中两个项的量纲是多少?

What are the dimensions of the two terms in the spring equation?

首先看一下简单的第二项 kx。它源于胡克定律,该定律指出,理想弹簧施加的力为 kx,其中 x 是弹簧相对于其平衡长度的伸长量。因此,第二项 kx 是一个力。那么第一项也是力吗?

Look first at the simple second term kx. It arises from Hooke’s law, which says that an ideal spring exerts a force kx where x is the extension of the spring relative to its equilibrium length. Thus the second term kx is a force. Is the first term also a force?

第一项 m(d²x / dt² )包含二阶导数 d²x / dt² 也就是我们熟悉的加速度。然而,许多微分方程包含我们不太熟悉的导数。流体力学的纳维-斯托克斯方程(第 2.4 节

The first term m(d2x/dt2) contains the second derivative d2x/dt2, which is familiar as an acceleration. Many differential equations, however, contain unfamiliar derivatives. The Navier–Stokes equations of fluid mechanics (Section 2.4),

图像

包含两个奇怪的导数: ( v · Δ ) vΔ 2 v。这些项的量纲是多少?

contain two strange derivatives: (v·Δ)v and Δ2v. What are the dimensions of those terms?

为了练习稍后处理这些复杂的项,现在让我们手动计算 d 2 x/dt 2的维度。由于 d 2 x/dt 2包含两个 2 的指数,x 是长度,t 是时间,因此 d 2 x/dt 2的维度可能为 L 2 T −2

To practice for later handling such complicated terms, let’s now find the dimensions of d2x/dt2 by hand. Because d2x/dt2 contains two exponents of 2, and x is length and t is time, d2x/dt2 might plausibly have dimensions of L2T−2.

图像 L 2 T −2 正确的尺寸吗?

Are L2T−2 the correct dimensions?

为了做出判断,我们可以使用1.3.2 节中的概念:微分符号 d 表示“一点点”。分子 d ² x 表示 dx 的 d,是“x 的一点点”。因此,它是一个长度。分母 dt ²可能表示 (dt) ²或 d(t ² )。[结果证明是 (dt) ²。] 无论哪种情况,它的维度都是 T ²。因此,二阶导数的维度为 LT −2

To decide, use the idea from Section 1.3.2 that the differential symbol d means “a little bit of.” The numerator d2x, meaning d of dx, is “a little bit of a little bit of x.” Thus, it is a length. The denominator dt2 could plausibly mean (dt)2 or d(t2). [It turns out to mean (dt)2.] In either case, its dimensions are T2. Therefore, the dimensions of the second derivative are LT−2:

图像

这种组合就是加速度,因此弹簧方程的第一项 m(d 2 x/dt 2 ) 是质量乘以加速度——使其具有与 kx 项相同的维度。

This combination is an acceleration, so the spring equation’s first term m(d2x/dt2) is mass times acceleration—giving it the same dimensions as the kx term.

问题 3.18 弹簧常数的量纲

Problem 3.18 Dimensions of spring constant

弹簧常数 k 的量纲是多少?

What are the dimensions of the spring constant k?

3.4.2 估计项的大小

3.4.2 Estimating the magnitudes of the terms

弹簧方程通过了量纲检验,因此值得分析以确定振动频率。方法是将每个项替换为其近似值。这些替换将把复杂的微分方程转化为简单的频率代数方程。

The spring equation passes the dimensions test, so it is worth analyzing to find the oscillation frequency. The method is to replace each term with its approximate magnitude. These replacements will turn a complicated differential equation into a simple algebraic equation for the frequency.

为了近似第一项 m(d 2 x/dt 2 ),使用显著变化近似法(第 3.3.3 节)来估计加速度 d 2 x/dt 2的大小。

To approximate the first term m(d2x/dt2), use the significant-change approximation (Section 3.3.3) to estimate the magnitude of the acceleration d2x/dt2.

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问题 3.19 解释指数

Problem 3.19 Explaining the exponents

分子只包含Δx的一次方,而分母包含Δt的二次方。这种差异怎么可能是正确的呢?

The numerator contains only the first power of Δx, whereas the denominator contains the second power of Δt. How can that discrepancy be correct?

要评估这个近似加速度,首先要确定一个显著的Δx,即什么能导致质量块位置发生显著变化。质量块在点 x = −x 0和 x = +x 0之间移动,因此位置的显著变化应该是峰峰值幅度 2x 0的显著比例。最简单的选择是Δx = x 0

To evaluate this approximate acceleration, first decide on a significant Δx—on what constitutes a significant change in the mass’s position. The mass moves between the points x = −x0 and x = +x0, so a significant change in position should be a significant fraction of the peak-to-peak amplitude 2x0. The simplest choice is Δx = x0.

现在估计 Δt:即物块移动相当于 Δx 的距离所需的时间。这个时间称为系统的特征时间,与振荡周期 T 有关。在一个周期内,物块来回移动,移动的距离为 4x 0 ,比 x 0远得多。如果 Δt 为 T/4 或 T/2π,那么在 Δt 时间内物块移动的距离将与 x 0相当。这些 Δt 的取值可以自然地解释为约为 1/ω,其中角频率 ω 与周期的关系定义:ω ≡ 2π/T。根据上述 Δx 和 Δt 的取值,m(d 2 x/dt 2 ) 项大约为 mx 0 ω 2

Now estimate Δt: the time for the block to move a distance comparable to Δx. This time—called the characteristic time of the system—is related to the oscillation period T. During one period, the mass moves back and forth and travels a distance 4x0—much farther than x0. If Δt were, say, T/4 or T/2π, then in the time Δt the mass would travel a distance comparable to x0. Those choices for Δt have a natural interpretation as being approximately 1/ω, where the angular frequency ω is connected to the period by the definition ω ≡ 2π/T. With the preceding choices for Δx and Δt, the m(d2x/dt2) term is roughly mx0ω2.

图像 “大致”是什么意思?

What does “is roughly” mean?

这句话的意思并非是 mx 0 ω 2与 m(d 2 x/dt 2 ) 的差值在 2 倍以内,因为 m(d 2 x/dt 2 ) 会变化,而 mx 02为常数。“大致”的意思是 m(d 2 x/dt 2 ) 的典型值或特征值(例如,其均方根值)与 mx 0 ω 2相当。让我们将这个含义包含在旋转符号 ∼ 中。那么,典型值估计值可以写成

The phrase cannot mean that mx0ω2 and m(d2x/dt2) are within, say, a factor of 2, because m(d2x/dt2) varies and mx02 is constant. Rather, “is roughly” means that a typical or characteristic magnitude of m(d2x/dt2)—for example, its root-mean-square value—is comparable to mx0ω2. Let’s include this meaning within the twiddle notation ∼. Then the typical-magnitude estimate can be written

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与“大致”的含义相同,即典型值是可比的,弹簧方程的第二项 kx 大致等于 kx 0。这两个项之和必须为零——这是弹簧方程的推论

With the same meaning of “is roughly”, namely that the typical magnitudes are comparable, the spring equation’s second term kx is roughly kx0. The two terms must add to zero—a consequence of the spring equation

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因此,这两个项的大小是可比较的:

Therefore, the magnitudes of the two terms are comparable:

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振幅 x 0除以 ω!当 x 0消失后,频率 ω 和振荡周期 T = 2π/ω 与振幅无关。[此推论使用了几种近似值,但这个结论是精确的(问题 3.20)。] 近似的角频率 ω 为图像

The amplitude x0 divides out! With x0 gone, the frequency ω and oscillation period T = 2π/ω are independent of amplitude. [This reasoning uses several approximations, but this conclusion is exact (Problem 3.20).] The approximated angular frequency ω is then

为了进行比较,根据问题 3.22可知,弹簧微分方程的精确解为:

For comparison, the exact solution of the spring differential equation is, from Problem 3.22,

图像

其中 ω 是图像近似的角频率也是精确的!

where ω is The approximated angular frequency is also exact!

问题 3.20 振幅独立性

Problem 3.20 Amplitude independence

使用量纲分析表明角频率 ω 不能依赖于振幅 x 0

Use dimensional analysis to show that the angular frequency ω cannot depend on the amplitude x0.

问题 3.21 检查所谓解决方案中的尺寸

Problem 3.21 Checking dimensions in the alleged solution

ωt 的维数是多少?cos ωt 的维数是多少?检查建议解 x = x 0 cos ωt 的维数,以及建议周期 2π 的维数。图像

What are the dimensions of ωt? What are the dimensions of cos ωt? Check the dimensions of the proposed solution x = x0 cos ωt, and the dimensions of the proposed period 2π

问题3.22 验证

Problem 3.22 Verification

证明 x = x 0 cos ωt ,ω =图像解弹簧微分方程

Show that x = x0 cos ωt with ω = solves the spring differential equation

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3.4.3 雷诺数的含义

3.4.3 Meaning of the Reynolds number

作为集中的另一个例子——特别是显著变化近似——让我们分析一下第 2.4 节中介绍的纳维-斯托克斯方程,

As a further example of lumping—in particular, of the significant-change approximation—let’s analyze the Navier–Stokes equations introduced in Section 2.4,

图像

并从中提取雷诺数rν/ν的物理意义。

and extract from them a physical meaning for the Reynolds number rν/ν.

为此,我们估计了惯性项 ( v · ) v和粘性项2 v的典型大小。

To do so, we estimate the typical magnitude of the inertial term (v·)v and of the viscous term 2v.

图像 惯性项的典型大小是多少?

What is the typical magnitude of the inertial term?

惯性项 ( v · ) ∇v包含空间导数∇v。根据显著变化近似(第 3.3.3 节),导数∇v大致等于

The inertial term (v·)∇v contains the spatial derivative ∇v. According to the significant-change approximation (Section 3.3.3), the derivative ∇v is roughly the ratio

图像

远离圆锥体的流速(空气速度)几乎为零,而靠近圆锥体(以速度 ν 运动)时,其速度与 ν 相当。因此,ν 或 ν 的合理比例会引起流速的显著变化。这种速度变化发生在与圆锥体大小相当的距离上:在几个圆锥体长度之外,空气几乎感觉不到圆锥体下落。因此∇ν ∼ ν/r。惯性项 ( ν · ) ν包含v的第二个因子,因此 ( v · ) v大约等于 v 2 /r。

The flow velocity (the velocity of the air) is nearly zero far from the cone and is comparable to ν near the cone (which is moving at speed ν). Therefore, ν, or a reasonable fraction of ν, constitutes a significant change in flow velocity. This speed change happens over a distance comparable to the size of the cone: Several cone lengths away, the air hardly knows about the falling cone. Thus ∇ν ∼ ν/r. The inertial term (ν·)ν contains a second factor of v, so (v·)v is roughly v2/r.

图像 粘性项的典型大小是多少?

What is the typical magnitude of the viscous term?

粘性项 v 2 v包含v的两个空间导数。由于每个空间导数都会对典型幅值贡献 1/r 的因子,因此 v∇ 2 v大约等于 vν/r 2。惯性项与粘性项的比值大约等于 (ν 2 /r)/(vν/r 2 )。该比值简化为 rν/v——我们熟悉的无量纲雷诺数。

The viscous term v2v contains two spatial derivatives of v. Because each spatial derivative contributes a factor of 1/r to the typical magnitude, v∇2v is roughly vν/r2. The ratio of the inertial term to the viscous term is then roughly (ν2/r)/(vν/r2). This ratio simplifies to rν/v—the familiar, dimensionless, Reynolds number.

因此,雷诺数衡量的是粘度的重要性。当 Re≫1 时,粘度项较小,粘度的影响可以忽略不计。它无法阻止邻近流体获得显著不同的速度,流动变为湍流。当 Re≫1 时图像,粘度项较大,粘度成为主要的物理效应。流动会像倒冷蜂蜜一样渗出。

Thus, the Reynolds number measures the importance of viscosity. When Re ≫ 1, the viscous term is small, and viscosity has a negligible effect. It cannot prevent nearby pieces of fluid from acquiring significantly different velocities, and the flow becomes turbulent. When Re 1, the viscous term is large, and viscosity is the dominant physical effect. The flow oozes, as when pouring cold honey.

3.5 预测单摆的周期

3.5 Predicting the period of a pendulum

集中不仅能将积分转化为乘法,还能将非线性微分方程转化为线性微分方程。我们以钟摆周期分析为例,几个世纪以来,钟摆一直是西方计时的基础。

Lumping not only turns integration into multiplication, it turns nonlinear into linear differential equations. Our example is the analysis of the period of a pendulum, for centuries the basis of Western timekeeping.

图像 钟摆的周期与其振幅有何关系?

How does the period of a pendulum depend on its amplitude?

振幅 θ 0是摆幅的最大角度;对于静止释放的无损耗摆,它也是释放角度。振幅的影响包含在摆微分方程的解中(方程的推导参见 [ 24 ]):

The amplitude θ0 is the maximum angle of the swing; for a lossless pendulum released from rest, it is also the angle of release. The effect of amplitude is contained in the solution to the pendulum differential equation (see [24] for the equation’s derivation):

图像

图像

分析将使用我们所有的工具:维度(第 3.5.2 节)、简单案例(第 3.5.1 节和第3.5.3节)和集中(第 3.5.4 节)。

The analysis will use all our tools: dimensions (Section 3.5.2), easy cases (Section 3.5.1 and Section 3.5.3), and lumping (Section 3.5.4).

问题 3.23 角度

Problem 3.23 Angles

解释为什么角度是无量纲的。

Explain why angles are dimensionless.

问题 3.24 检查和使用尺寸

Problem 3.24 Checking and using dimensions

摆锤方程的量纲正确吗?使用量纲分析来证明该方程不能包含摆锤的质量(除非将其作为公因数除以零)。

Does the pendulum equation have correct dimensions? Use dimensional analysis to show that the equation cannot contain the mass of the bob (except as a common factor that divides out).

3.5.1 小振幅:应用极端情况

3.5.1 Small amplitudes: Applying extreme cases

钟摆方程很难计算,因为它包含一个非线性因子 sin θ。幸运的是,在小振幅极端情况 θ 0 下,该因子很容易计算。在该极限下,三角形的高 sin θ 几乎等于弧长 θ。因此,对于小角度,sin θ θ。

The pendulum equation is difficult because of its nonlinear factor sin θ. Fortunately, the factor is easy in the small-amplitude extreme case θ 0. In that limit, the height of the triangle, which is sin θ, is almost exactly the arclength θ. Therefore, for small angles, sin θ θ.

图像

问题 3.25 弦近似

Problem 3.25 Chord approximation

sin θ ≈ θ 近似值用一条垂直的直线代替了圆弧。为了得到更精确的近似值,可以用弦(一条非垂直的直线)代替圆弧。sin θ 的近似值是多少?

The sin θ ≈ θ approximation replaces the arc with a straight, vertical line. To make a more accurate approximation, replace the arc with the chord (a straight but nonvertical line). What is the resulting approximation for sin θ?

在小振幅极端情况下,钟摆方程变为线性:

In the small-amplitude extreme, the pendulum equation becomes linear:

图像

将此方程与弹簧质量方程(第 3.4 节)进行比较

Compare this equation to the spring–mass equation (Section 3.4)

图像

方程中,x 类似于 θ,k/m 类似于 g/l。弹簧质量系统的频率为 ω = 图像,周期为 T = 2π/ω = 图像。对于单摆方程,相应的周期为

The equations correspond with x analogous to θ and k/m analogous to g/l. The frequency of the spring–mass system is ω = , and its period is T = 2π/ω = . For the pendulum equation, the corresponding period is

图像

(本分析是对类比方法的预览,类比方法是第六章的主题。)

(This analysis is a preview of the method of analogy, which is the subject of Chapter 6.)

问题 3.26 检查尺寸

Problem 3.26 Checking dimensions

句号的图像尺寸是否正确?

Does the period have correct dimensions?

问题 3.27 检查极端情况

Problem 3.27 Checking extreme cases

在极端情况 g → ∞ 和 g → 0下周期 T = 是否图像有意义?

Does the period T = make sense in the extreme cases g → ∞ and g → 0?

问题 3.28 可能的巧合

Problem 3.28 Possible coincidence

g ≈ π 2 ms −2是巧合吗?(有关钟摆的详细历史讨论,请参见 [ 1 ] 以及更广泛的文献 [ 4 , 27 , 42 ]。)

Is it a coincidence that g ≈ π2 ms−2? (For an extensive historical discussion that involves the pendulum, see [1] and more broadly also [4, 27, 42].)

问题 3.29 圆锥摆的常数

Problem 3.29 Conical pendulum for the constant

无量纲因子 2π 可以运用惠更斯 [ 15 , p. 79 ]的一个洞见推导出来:分析一个在水平圆周运动的单摆(圆锥摆)的运动。将其二维运动投影到垂直屏幕上,会产生一维单摆运动,因此二维运动的周期与一维单摆运动的周期相同!运用这一思想以及牛顿运动定律来解释 2π。

The dimensionless factor of 2π can be derived using an insight from Huygens [15, p. 79]: to analyze the motion of a pendulum moving in a horizontal circle (a conical pendulum). Projecting its two-dimensional motion onto a vertical screen produces one-dimensional pendulum motion, so the period of the two-dimensional motion is the same as the period of one-dimensional pendulum motion! Use that idea along with Newton’s laws of motion to explain the 2π.

图像

3.5.2 任意振幅:应用量纲分析

3.5.2 Arbitrary amplitudes: Applying dimensional analysis

如果振幅 θ 0不再小,上述结果可能会发生变化。

The preceding results might change if the amplitude θ0 is no longer small.

图像 随着θ 0 的 增加,周期是增加、保持不变还是减少?

As θ0 increases, does the period increase, remain constant, or decrease?

如果使用无量纲群(第 2.4.1 节)来表达,任何分析都会变得更加清晰。本问题涉及周期 T、长度 l、引力强度 g 和振幅 θ 0。因此,T 可以属于无量纲群图像。由于角度是无量纲的,所以 θ 0本身也是一个群。这两个群图像和 θ 0相互独立,可以完整地描述该问题(问题 3.30)。

Any analysis becomes cleaner if expressed using dimensionless groups (Section 2.4.1). This problem involves the period T, length l, gravitational strength g, and amplitude θ0. Therefore, T can belong to the dimensionless group . Because angles are dimensionless,θ0 is itself a group. The two groups and θ0 are independent and fully describe the problem (Problem 3.30).

一个有益的对比是理想的弹簧质量系统。周期 T、弹簧常数 k 和质量 m 可以构成无量纲组图像;但振幅 x 0作为唯一包含长度的量,不能成为任何无量纲组的一部分(习题 3.20),因此不会影响弹簧质量系统的周期。相比之下,单摆的振幅θ 0已经是一个无量纲群,因此它可以影响系统的周期。

An instructive contrast is the ideal spring–mass system. The period T, spring constant k, and mass m can form the dimensionless group ; but the amplitude x0, as the only quantity containing a length, cannot be part of any dimensionless group (Problem 3.20) and cannot therefore affect the period of the spring–mass system. In contrast, the pendulum’s amplitude θ0 is already a dimensionless group, so it can affect the period of the system.

图像

问题 3.30 选择无量纲组

Problem 3.30 Choosing dimensionless groups

检查周期 T、长度 l、引力强度 g 和振幅 θ 0是否构成两个独立的无量纲组。在构建用于分析周期的组时,为什么 T 只能出现在一个组中?为什么 θ 0不能与 T 出现在同一个组中?

Check that period T, length l, gravitational strength g, and amplitude θ0 produce two independent dimensionless groups. In constructing useful groups for analyzing the period, why should T appear in only one group? And why should θ0 not appear in the same group as T?

两个无量纲群产生一般无量纲形式

Two dimensionless groups produce the general dimensionless form

图像

所以

so

图像

因为图像当θ 0 = 0(小振幅极限)时,将2π分解出来以简化后续方程,并定义无量纲周期h如下:

Because when θ0 = 0 (the small-amplitude limit), factor out the 2π to simplify the subsequent equations, and define a dimensionless period h as follows:

图像

函数 h 包含了振幅如何影响单摆周期的所有信息。利用 h,关于周期的原始问题变成了:h 是振幅的增函数、常数函数还是减函数?下一节将解答这个问题。

The function h contains all information about how amplitude affects the period of a pendulum. Using h, the original question about the period becomes the following: Is h an increasing, constant, or decreasing function of amplitude? This question is answered in the following section.

3.5.3 大振幅:再次出现极端情况

3.5.3 Large amplitudes: Extreme cases again

为了猜测 h 作为振幅函数的一般行为,有用的线索来自于在两个振幅下评估 h。一个容易的振幅是零振幅的极值,其中 h(0) = 1。第二个容易的振幅是大振幅的另一个极值。

For guessing the general behavior of h as a function of amplitude, useful clues come from evaluating h at two amplitudes. One easy amplitude is the extreme of zero amplitude, where h(0) = 1. A second easy amplitude is the opposite extreme of large amplitudes.

图像 在大振幅下,周期如何变化?作为这个问题的一部分,什么是大振幅?

How does the period behave at large amplitudes? As part of that question, what is a large amplitude?

一个有趣的大振幅是π/2,这意味着将摆锤从水平方向释放。然而,在π/2处,精确的h是以下可怕的表达式(问题3.31):

An interesting large amplitude is π/2, which means releasing the pendulum from horizontal. However, at π/2 the exact h is the following awful expression (Problem 3.31):

图像

这个积分小于、等于还是大于 1?谁知道呢?这个积分很可能没有封闭形式,需要数值计算(问题 3.32)。

Is this integral less than, equal to, or more than 1? Who knows? The integral is likely to have no closed form and to require numerical evaluation (Problem 3.32).

问题 3.31 h的一般表达式

Problem 3.31 General expression for h

利用能量守恒来证明周期是

Use conservation of energy to show that the period is

图像

确认等效无量纲语句是

Confirm that the equivalent dimensionless statement is

图像

对于水平释放,θ 0 = π/2,并且

For horizontal release, θ0 = π/2, and

图像

问题 3.32 水平释放的数值评估

Problem 3.32 Numerical evaluation for horizontal release

为什么集中方法(第 3.2 节)对问题 3.31中的积分不起作用?使用数值积分计算 h(π/2)。

Why do the lumping recipes (Section 3.2) fail for the integrals in Problem 3.31? Compute h(π/2) using numerical integration.

因为 θ 0 = π/2 不是一个有用的极值,所以可以取更极端的值。试试 θ 0 = π,这意味着将摆锤从垂直位置释放。然而,如果摆锤用绳子连接到支点,垂直释放将意味着摆锤垂直下落而不是振荡。这种新奇的行为既不包含在摆锤微分方程中,也无法用摆锤微分方程描述。

Because θ0 = π/2 is not a helpful extreme, be even more extreme. Try θ0 = π, which means releasing the pendulum bob from vertical. If the bob is connected to the pivot point by a string, however, a vertical release would mean that the bob falls straight down instead of oscillating. This novel behavior is neither included in nor described by the pendulum differential equation.

图像

幸运的是,改进一个思想实验并不费力:用一根无质量的钢棒代替绳子。在θ 0 = π时,摆锤完全平衡,永远倒挂着,因此T(π) = ∞,h(π) = ∞。因此,h(π) > 1,h(0) = 1。根据这些数据,最有可能的推测是h随振幅单调递增。虽然h可能先减小后增大,但对于这样一个纯粹的微分方程来说,这种曲折的行为会令人感到意外。(关于h在θ 0 = π附近的行为,参见习题3.34)。

Fortunately, a thought experiment is cheap to improve: Replace the string with a massless steel rod. Balanced perfectly at θ0 = π, the pendulum bob hangs upside down forever, so T(π) = ∞ and h(π) = ∞. Thus, h(π) > 1 and h(0) = 1. From these data, the most likely conjecture is that h increases monotonically with amplitude. Although h could first decrease and then increase, such twists and turns would be surprising behavior from such a clean differential equation. (For the behavior of h near θ0 = π, see Problem 3.34).

问题 3.33 小但非零的振幅

Problem 3.33 Small but nonzero amplitude

当振幅接近π时,无量纲周期h趋于无穷大;在零振幅时,h = 1。但是h的导数呢?在零振幅(θ 0 = 0)时,h(θ 0 )的斜率为零(曲线A)还是正斜率(曲线B)?

As the amplitude approaches π, the dimensionless period h diverges to infinity; at zero amplitude, h = 1. But what about the derivative of h? At zero amplitude (θ0 = 0), does h(θ0) have zero slope (curve A) or positive slope (curve B)?

图片

问题 3.34 近乎垂直的释放

Problem 3.34 Nearly vertical release

想象一下,将钟摆从几乎垂直的位置释放:初始角度为π − β,β 很小。作为β的函数,钟摆旋转一个显著角度(比如1弧度)大约需要多长时间?利用这些信息预测当θ 0 ≈ π时h(θ 0 )的变化。使用表格中的值检查并改进你的猜想。然后预测h(π − 10 −5 )。

Imagine releasing the pendulum from almost vertical: an initial angle π − β with β tiny. As a function of β, roughly how long does the pendulum take to rotate by a significant angle—say, by 1 rad? Use that information to predict how h(θ0) behaves when θ0 ≈ π. Check and refine your conjectures using the tabulated values. Then predict h(π − 10−5).

图片

3.5.4 中等振幅:应用集中法

3.5.4 Moderate amplitudes: Applying lumping

h单调递增的猜想是利用零和垂直振幅的极值推导出来的,因此它应该适用于中间振幅。在轻信这一说法之前,先回顾一下军控谈判中的一句谚语:“信任,但要核实。”

The conjecture that h increases monotonically was derived using the extremes of zero and vertical amplitude, so it should apply at intermediate amplitudes. Before taking that statement on faith, recall a proverb from arms-control negotiations: “Trust, but verify.”

图像 在中等(小但不为零)振幅下,周期或其无量纲h是否会随着振幅的增加而增加?

At moderate (small but nonzero) amplitudes, does the period, or its dimensionless cousin h, increase with amplitude?

在零振幅极值时,sinθ接近于θ。该近似将非线性摆方程

In the zero-amplitude extreme, sin θ is close to θ. That approximation turned the nonlinear pendulum equation

图片

进入线性理想弹簧方程——其中周期与振幅无关。

into the linear, ideal-spring equation—in which the period is independent of amplitude.

然而,在非零振幅时,θ 和 sin θ 不同,并且它们的差会影响周期。为了解释这个差异并预测周期,将 sin θ 分解为可处理因子 θ 和调整因子 f(θ)。所得方程为

At nonzero amplitude, however, θ and sin θ differ and their difference affects the period. To account for the difference and predict the period, split sin θ into the tractable factor θ and an adjustment factor f(θ). The resulting equation is

图片

非常量 f(θ) 概括了摆方程的非线性。当 很小时,f(θ) ≈ 1:摆的行为类似于线性理想弹簧系统。但当 θ 较大时,f(θ) 会显著小于 1,使得理想弹簧近似值显著不准确。变化过程通常难以分析——例如,参见习题 3.31中那些糟糕的积分。为了解决这个问题,可以采用集中近似法,将变化的 f(θ) 替换为常数。

The nonconstant f(θ) encapsulates the nonlinearity of the pendulum equation. When is tiny, f(θ) ≈ 1: The pendulum behaves like a linear, ideal-spring system. But when θ is large, f(θ) falls significantly below 1, making the ideal-spring approximation significantly inaccurate. As is often the case, a changing process is difficult to analyze—for example, see the awful integrals in Problem 3.31. As a countermeasure, make a lumping approximation by replacing the changing f(θ) with a constant.

图片

最简单的常数是 f(0)。则摆的微分方程变为

The simplest constant is f(0). Then the pendulum differential equation becomes

图片

图片

这个方程同样是理想弹簧方程。在这个近似中,周期与振幅无关,因此对于所有振幅,h = 1。为了确定未近似摆的周期如何依赖于振幅,f(θ) → f(0) 的集中近似会丢弃太多信息。

This equation is, again, the ideal-spring equation. In this approximation, period does not depend on amplitude, so h = 1 for all amplitudes. For determining how the period of an unapproximated pendulum depends on amplitude, the f(θ) → f(0) lumping approximation discards too much information.

因此,用另一个极值 f(θ 0 ) 代替 f(θ),则单摆方程变为

Therefore, replace f(θ) with the other extreme f(θ0). Then the pendulum equation becomes

图片

图片

图像 这个方程是线性的吗?它描述的是什么物理系统?

Is this equation linear? What physical system does it describe?

因为 f(θ 0 ) 是一个常数,所以这个方程是线性的!它描述的是一个零振幅摆,摆位于一个重力 g eff略弱于地球重力的行星上——如下图所示:

Because f(θ0) is a constant, this equation is linear! It describes a zero-amplitude pendulum on a planet with gravity geff that is slightly weaker than earth gravity—as shown by the following slight regrouping:

图片

因为零振幅摆的周期为图片,所以零振幅、低重力摆的周期为

Because the zero-amplitude pendulum has period , the zero-amplitude, low-gravity pendulum has period

图片

使用无量纲周期 h 可以避免写出 2π、l 和 g 的因子,从而得到简单的预测

Using the dimensionless period h avoids writing the factors of 2π, l, and g, and it yields the simple prediction

图片

图片

在中等振幅下,近似值与精确的无量纲周期(深色曲线)非常接近。此外,它还预测了 h(π) = ∞,因此它与将钟摆从直立状态释放的思想实验(第 3.5.3 节)相符。

At moderate amplitudes the approximation closely follows the exact dimensionless period (dark curve). As a bonus, it also predicts h(π) = ∞, so it agrees with the thought experiment of releasing the pendulum from upright (Section 3.5.3).

图像 10°振幅时的周期比零振幅时的周期大多少

How much larger than the period at zero amplitude is the period at 10° amplitude?

10° 的振幅大约等于 0.17 弧度,这是一个中等角度,因此 h 的近似预测本身可以用泰勒级数精确地近似。sin 的泰勒级数起始于 θ−θ 3 /6,因此

A 10° amplitude is roughly 0.17 rad, a moderate angle, so the approximate prediction for h can itself accurately be approximated using a Taylor series. The Taylor series for sin begins θ−θ3/6, so

图片

那么 h(θ 0 ),大致等于 f(θ 0 ) −1/2,变为

Then h(θ0), which is roughly f(θ0)−1/2, becomes

图片

另一个泰勒级数可得 (1 + x) −1/2 ≈ 1 − x/2(对于较小的 x)。因此,

Another Taylor series yields (1 + x)−1/2 ≈ 1 − x/2 (for small x). Therefore,

图片

恢复尺寸量即可得出周期本身。

Restoring the dimensioned quantities gives the period itself.

图片

与零振幅时的周期相比,10°振幅会使周期略微增加,约为θ² / 0 /12≈0.0025,即0.25%。即使在中等振幅下,周期也几乎与振幅无关!

Compared to the period at zero amplitude, a 10° amplitude produces a fractional increase of roughly θ20/12 ≈ 0.0025 or 0.25%. Even at moderate amplitudes, the period is nearly independent of amplitude!

问题 3.35 重新审视斜率

Problem 3.35 Slope revisited

使用前面 h(θ 0 )的结果来检验问题 3.33中关于 θ 0 = 0 时 h(θ 0 )斜率的结论。

Use the preceding result for h(θ0) to check your conclusion in Problem 3.33 about the slope of h(θ0) at θ0 = 0.

图像我们的集中近似是否低估或高估了周期?

Does our lumping approximation underestimate or overestimate the period?

集中近似法通过将 f(θ) 替换为 f(θ 0 )来简化摆的微分方程。等效地,它假设质量始终保持在运动的端点,即 |θ| = θ 0。事实上,摆的大部分时间都处于中间位置,即 |θ| < θ 0且 f(θ) > f(θ 0 )。因此,平均 f 大于 f(θ 0 )。由于 h 与 f 成反比(h = f −1/2),因此 f(θ) f(θ 0 ) 集中近似法会高估 h 和周期。

The lumping approximation simplified the pendulum differential equation by replacing f(θ) with f(θ0). Equivalently, it assumed that the mass always remained at the endpoints of the motion where |θ| = θ0. Instead, the pendulum spends much of its time at intermediate positions where |θ| < θ0 and f(θ) > f(θ0). Therefore, the average f is greater than f(θ0). Because h is inversely related to f (h = f−1/2), the f(θ) f(θ0) lumping approximation overestimates h and the period.

f(θ) f(0) 的集中近似预测了 ,但图片低估了周期。因此,图片周期近似中 项的真实系数

The f(θ) f(0) lumping approximation, which predicts , underestimates the period. Therefore, the true coefficient of the term in the period approximation

图片

介于0和1/12之间。一个自然的猜测是,该系数位于这两个极值之间的中间值,即1/24。然而,钟摆在极值点(f(θ) = f(θ 0 ))处花费的时间,比在平衡位置(f(θ) = f(0))附近花费的时间要多。因此,真实系数可能更接近1/12(f(θ) → f(θ 0 )近似值的预测值),而不是0。一个更合理的猜测可能是0到1/12之间三分之二的值,即1/18。

lies between 0 and 1/12. A natural guess is that the coefficient lies halfway between these extremes—namely, 1/24. However, the pendulum spends more time toward the extremes (where f(θ) = f(θ0)) than it spends near the equilibrium position (where f(θ) = f(0)). Therefore, the true coefficient is probably closer to 1/12—the prediction of the f(θ) → f(θ0) approximation—than it is to 0. An improved guess might be two-thirds of the way from 0 to 1/12, namely 1/18.

相比之下,单摆微分方程的完整逐次逼近解给出了以下周期 [ 13 , 33 ]:

In comparison, a full successive-approximation solution of the pendulum differential equation gives the following period [13, 33]:

图片

我们根据经验猜测的 1/18 与真实系数 1/16 非常接近!

Our educated guess of 1/18 is very close to the true coefficient of 1/16!

3.6 总结和进一步的问题

3.6 Summary and further problems

集总法彻底颠覆了微积分。微积分通过将变化过程划分成越来越细的区间来分析它,而集总法则通过将变化过程合并成一个不变的过程来简化它。它把曲线变成直线,把复杂的积分变成乘法,把轻微的非线性微分方程变成线性微分方程。

Lumping turns calculus on its head. Whereas calculus analyzes a changing process by dividing it into ever finer intervals, lumping simplifies a changing process by combining it into one unchanging process. It turns curves into straight lines, difficult integrals into multiplication, and mildly nonlinear differential equations into linear differential equations.

……弯曲之处必变为正直,崎岖之处必变为平坦。(以赛亚书 40:4)

. . . the crooked shall be made straight, and the rough places plain. (Isaiah 40:4)

问题 3.36 另一个衰减函数的 FWHM

Problem 3.36 FWHM for another decaying function

使用 FWHM 启发式方法来估计

Use the FWHM heuristic to estimate

图片

然后将估计值与 的精确值进行比较图片。这是一个有趣的附加问题,推导出精确值。

Then compare the estimate with the exact value of . For an enjoyable additional problem, derive the exact value.

问题 3.37 假想摆方程

Problem 3.37 Hypothetical pendulum equation

假设钟摆方程是

Suppose the pendulum equation had been

图片

周期 T 与振幅 θ 0有何关系?具体来说,随着 θ 0 的增加,T 会减小、保持不变还是增大?在振幅为零时,斜率 dT/dθ 0是多少?将你的结果与问题 3.33的结果进行比较。

How would the period T depend on amplitude θ0? In particular, as θ0 increases, would T decrease, remain constant, or increase? What is the slope dT/dθ0 at zero amplitude? Compare your results with the results of Problem 3.33.

对于较小但非零的 θ 0 ,找到无量纲周期 h(θ 0 )的近似表达式,并用它来检查您之前的结论。

For small but nonzero θ0, find an approximate expression for the dimensionless period h(θ0) and use it to check your previous conclusions.

问题 3.38 高斯1 -sigma 尾部

Problem 3.38 Gaussian 1-sigma tail

均值为零、方差为单位的高斯概率密度函数为

The Gaussian probability density function with zero mean and unit variance is

图片

尾部面积是统计学中的一个重要量,但它没有封闭的形式。在这个问题中,你需要估计1-sigma尾部的面积

The area of its tail is an important quantity in statistics, but it has no closed form. In this problem you estimate the area of the 1-sigma tail

图片

a. 画出上面的高斯分布图,并对 1-sigma 尾部进行阴影处理。

a. Sketch the above Gaussian and shade the 1-sigma tail.

b. 使用 1/e 集中启发式方法(第 3.2.1 节)来估计面积。

b. Use the 1/e lumping heuristic (Section 3.2.1) to estimate the area.

c. 使用 FWHM 启发式方法来估计面积。

c. Use the FWHM heuristic to estimate the area.

d. 将两个集中估计与数值积分的结果进行比较:

d. Compare the two lumping estimates with the result of numerical integration:

图片

其中 erf(z) 是误差函数。

where erf(z) is the error function.

问题 3.39 远距高斯尾部

Problem 3.39 Distant Gaussian tails

对于典型高斯概率,估计其 n-sigma 尾部的面积(n 较大时)。换句话说,估计

For the canonical probability Gaussian, estimate the area of its n-sigma tail (for large n). In other words, estimate

图片

4

4

图片样张

Pictorial proofs

你是否曾经尝试过一个证明,理解并确认了每一步,却仍然不相信这个定理?你意识到这个定理是正确的,但却不知道它为什么是正确的。

Have you ever worked through a proof, understood and confirmed each step, yet still not believed the theorem? You realize that the theorem is true, but not why it is true.

为了在一个熟悉的例子中看到同样的对比,想象一下,当你得知你的孩子发烧时,听到的温度是华氏度还是摄氏度,以你不太熟悉的为准。在我的日常生活中,温度大多以华氏度为单位。因此,当我听到40°C时,我的反应分为两个阶段:

To see the same contrast in a familiar example, imagine learning that your child has a fever and hearing the temperature in Fahrenheit or Celsius degrees, whichever is less familiar. In my everyday experience, temperatures are mostly in Fahrenheit. When I hear about a temperature of 40°C, I therefore react in two stages:

  1. 我将 40° C 转换为华氏度:40 × 1.8 + 32 = 104。
  2. I convert 40° C to Fahrenheit: 40 × 1.8 + 32 = 104.
  3. 我反应过来:“哇,104°F。太危险了!赶紧去看医生!”
  4. I react: “Wow, 104° F. That’s dangerous! Get thee to a doctor!”

摄氏温度虽然在象征意义上与华氏温度相当,但却没有引起任何反应。只有当温度转换将温度与我的经历联系起来时,我的危险感知才会被激活。

The Celsius temperature, although symbolically equivalent to the Fahrenheit temperature, elicits no reaction. My danger sense activates only after the temperature conversion connects the temperature to my experience.

无论是证明还是陌生的温度,符号化的描述都比能够触动我们感知系统的论证更缺乏说服力。原因在于我们的大脑是如何获得符号推理能力的。(参见《进化中的大脑》 [ 2 ],了解大脑的图解式学术史。)符号化、序列化的推理需要语言,而语言已经进化仅用了10⁻。虽然10⁻跨越了许多人的生命,但对于进化而言,它不过是转瞬即逝的瞬间。尤其与人类感知硬件进化的时间跨度相比,这显得格外短暂:几亿年来,生物体不断完善着听觉、嗅觉、味觉、触觉和视觉的能力。

A symbolic description, whether a proof or an unfamiliar temperature, is unconvincing compared to an argument that speaks to our perceptual system. The reason lies in how our brains acquired the capacity for symbolic reasoning. (See Evolving Brains [2] for an illustrated, scholarly history of the brain.) Symbolic, sequential reasoning requires language, which has evolved for only 105 yr. Although 105 yr spans many human lifetimes, it is an evolutionary eyeblink. In particular, it is short compared to the time span over which our perceptual hardware has evolved: For several hundred million years, organisms have refined their capacities for hearing, smelling, tasting, touching, and seeing.

进化在我们感知能力上的投入比在符号推理能力上的投入要长1000倍。与感知硬件相比,我们的符号序列硬件发展得比较晚,发展滞后。毫不奇怪,我们的感知能力远远超过符号能力。即使是像下国际象棋大师级比赛这样看似高水平的符号活动,也主要依靠感知硬件[ 16 ]。看到一个想法能给我们带来深刻的理解,而符号描述很难与之匹敌。

Evolution has worked 1000 times longer on our perceptual abilities than on our symbolic-reasoning abilities. Compared to our perceptual hardware, our symbolic, sequential hardware is an ill-developed latecomer. Not surprisingly, our perceptual abilities far surpass our symbolic abilities. Even an apparently high-level symbolic activity such as playing grandmaster chess uses mostly perceptual hardware [16]. Seeing an idea conveys to us a depth of understanding that a symbolic description of it cannot easily match.

问题 4.1 计算机与人类

Problem 4.1 Computers versus people

在诸如扩展 (x + 2y) 50 之类的任务上,计算机比人类快得多。在诸如识别人脸或气味之类的任务上,即使是幼儿也比现在的计算机快得多。你如何解释这种差异?

At tasks like expanding (x + 2y)50, computers are much faster than people. At tasks like recognizing faces or smells, even young children are much faster than current computers. How do you explain these contrasts?

问题 4.2 感知重要性的语言证据

Problem 4.2 Linguistic evidence for the importance of perception

用你最喜欢的语言,想想理解的许多感官同义词(例如,掌握)。

In your favorite language(s), think of the many sensory synonyms for understanding (for example, grasping).

4.1 奇数加法

4.1 Adding odd numbers

为了说明图片的价值,让我们求出前 n 个奇数的和(也是问题 2.25的主题):

To illustrate the value of pictures, let’s find the sum of the first n odd numbers (also the subject of Problem 2.25):

图像

简单的情况,例如 n = 1、2 或 3,可以得出 S n = n 2的猜想。但是如何证明这个猜想呢?标准的符号证明方法是归纳法:

Easy cases such as n = 1, 2, or 3 lead to the conjecture that Sn = n2. But how can the conjecture be proved? The standard symbolic method is proof by induction:

  1. 验证对于基本情况n = 1,S n = n 2。在这种情况下,S 1为 1,n 2也为1 ,因此基本情况得到验证。
  2. Verify that Sn = n2 for the base case n = 1. In that case, S1 is 1, as is n2, so the base case is verified.
  3. 提出归纳假设:假设当 m 小于或等于最大值 n 时, S m = m 2。对于此证明,以下较弱的归纳假设就足够了:

    图像

    换句话说,我们仅在 m = n 的情况下才假设该定理。

  4. Make the induction hypothesis: Assume that Sm = m2 for m less than or equal to a maximum value n. For this proof, the following, weaker induction hypothesis is sufficient:

    In other words, we assume the theorem only in the case that m = n.

  5. 执行归纳步骤:利用归纳假设证明 S n+1 = (n + 1) 2。和 S n+1分为两部分:

    图像

    根据归纳假设,右边的和是 n 2。因此

    图像

    即(n+1) 2;从而证明定理。

  6. Perform the induction step: Use the induction hypothesis to show that Sn+1 = (n + 1)2. The sum Sn+1 splits into two pieces:

    Thanks to the induction hypothesis, the sum on the right is n2. Thus

    which is (n + 1)2; and the theorem is proved.

尽管这些步骤证明了该定理,但为什么和S n最终会变成 n 2仍然难以捉摸。

Although these steps prove the theorem, why the sum Sn ends up as n2 still feels elusive.

这种缺失的理解——韦特海默[ 48 ]所描述的那种格式塔式洞见——需要用图形来证明。首先,把每个奇数画成一个L形的拼图:

That missing understanding—the kind of gestalt insight described by Wertheimer [48]—requires a pictorial proof. Start by drawing each odd number as an L-shaped puzzle piece:

图像

图像 这些部分如何组合在一起?

How do these pieces fit together?

然后通过将拼图碎片拼凑在一起来计算 S n ,如下所示:

Then compute Sn by fitting together the puzzle pieces as follows:

图像

每个连续的奇数——每一块——都会使正方形的高度和宽度各增加 1 个单位,因此这 n 项构成了一个 n × n 的正方形。[或者它是一个 (n − 1) × (n − 1) 的正方形?] 因此,它们的和是 n² 掌握了这个图解证明之后,你就不会忘记为什么将前 n 个奇数相加会得到

Each successive odd number—each piece—extends the square by 1 unit in height and width, so the n terms build an n × n square. [Or is it an (n − 1) × (n − 1) square?] Therefore, their sum is n2. After grasping this pictorial proof, you cannot forget why adding up the first n odd numbers produces n2.

问题 4.3 三角数

Problem 4.3 Triangular numbers

画一幅或多幅图来表明

Draw a picture or pictures to show that

图像

然后证明

Then show that

图像

问题 4.4 三维

Problem 4.4 Three dimensions

画一张图来表明

Draw a picture to show that

图像

用图形解释加数 3k 2 + 3k + 1 中的 1; 3k 2中的3 和 k 2;以及 3k 中的 3 和 k。

Give pictorial explanations for the 1 in the summand 3k2 + 3k + 1; for the 3 and the k2 in 3k2; and for the 3 and the k in 3k.

4.2 算术平均数和几何平均数

4.2 Arithmetic and geometric means

下一个图形证明从两个非负数(例如 3 和 4)开始,并比较以下两个平均值:

The next pictorial proof starts with two nonnegative numbers—for example, 3 and 4—and compares the following two averages:

图像

图像

尝试另一对数字,例如 1 和 2。算术平均值是 1.5,几何平均值是图像。对于这两对数字,几何平均值都小于算术平均值。这种模式很普遍;这就是著名的算术平均值-几何平均值(AM-GM)不等式 [ 18 ]:

Try another pair of numbers—for example, 1 and 2. The arithmetic mean is 1.5; the geometric mean is . For both pairs, the geometric mean is smaller than the arithmetic mean. This pattern is general; it is the famous arithmetic-mean–geometric-mean (AM–GM) inequality [18]:

图像

(不等式要求 a, b 图像0。)

(The inequality requires that a, b 0.)

问题 4.5 更多数值例子

Problem 4.5 More numerical examples

使用不同的数值例子检验AM-GM不等式。当a和b彼此接近时,你注意到了什么?你能形式化地描述这种模式吗?(另见问题4.16。)

Test the AM–GM inequality using varied numerical examples. What do you notice when a and b are close to each other? Can you formalize the pattern? (See also Problem 4.16.)

4.2.1 符号证明

4.2.1 Symbolic proof

AM-GM不等式有图示证明和符号证明。符号证明以(a − b) 2开头——这是一个令人惊讶的选择,因为不等式包含a + b,而不是a − b。第二个奇怪的选择是构造(a − b) 2。它是非负的,所以a 2 − 2ab + b 2 图像 0。现在,神奇地决定在两边都加上4ab。结果是

The AM–GM inequality has a pictorial and a symbolic proof. The symbolic proof begins with (a − b)2—a surprising choice because the inequality contains a + b rather than a − b. The second odd choice is to form (a − b)2. It is nonnegative, so a2 − 2ab + b2 0. Now magically decide to add 4ab to both sides. The result is

图像

左边是 (a + b) 2,所以 a + b 图像2图像

The left side is (a + b)2, so a + b 2 and

图像

虽然每一步都很简单,但整个链条却像魔术一样,留下了“为什么”的谜团。如果代数运算以“为什么”结束,图像它看起来就不会那么明显错误了。相反,一个令人信服的证明会让我们觉得这个不等式必然成立。

Although each step is simple, the whole chain seems like magic and leaves the why mysterious. If the algebra had ended with it would not look obviously wrong. In contrast, a convincing proof would leave us feeling that the inequality cannot help but be true.

4.2.2 图片证明

4.2.2 Pictorial proof

这种满足感是通过图形证明来实现的。

This satisfaction is provided by a pictorial proof.

图像 几何平均数在图形或几何上有何表现?

What is pictorial, or geometric, about the geometric mean?

几何平均值的几何图形始于一个直角三角形。将其斜边水平放置;然后以高 x 为界,将其分成明暗两个子三角形。斜边分成两个长度 a 和 b,高 x 为它们的几何平均值图像

A geometric picture for the geometric mean starts with a right triangle. Lay it with its hypotenuse horizontal; then cut it with the altitude x into the light and dark subtriangles. The hypotenuse splits into two lengths a and b, and the altitude x is their geometric mean .

图像

图像 为什么海拔 x 等于 图像

Why is the altitude x equal to

为了证明图像,将小三角形旋转并放置在大三角形上,比较小的深色三角形和大的浅色三角形。这两个三角形相似!因此,它们的长宽比(短边与长边的比值)相同。用符号表示为 x/a = b/x:因此,高 x 是几何平均数图像

To show that , compare the small, dark triangle to the large, light triangle by rotating the small triangle and laying it on the large triangle. The two triangles are similar! Therefore, their aspect ratios (the ratio of the short to the long side) are identical. In symbols, x/a = b/x: The altitude x is therefore the geometric mean .

图像

未切割的直角三角形代表 AM-GM 不等式的几何平均部分。算术平均 (a + b)/2 也表示为斜边的一半。因此,该不等式表明

The uncut right triangle represents the geometric-mean portion of the AM–GM inequality. The arithmetic mean (a + b)/2 also has a picture, as one-half of the hypotenuse. Thus, the inequality claims that

图像

可惜的是,这种说法并不明显。

Alas, this claim is not pictorially obvious.

图像 你能找到算术平均值的另一种几何解释,使 AM-GM 不等式在图形上显而易见吗?

Can you find an alternative geometric interpretation of the arithmetic mean that makes the AM–GM inequality pictorially obvious?

算术平均值也是直径为 a + b 的圆的半径。因此,在三角形外接一个半圆,使圆的直径等于斜边 a + b(问题 4.7)。高不能超过半径;因此,

The arithmetic mean is also the radius of a circle with diameter a + b. Therefore, circumscribe a semicircle around the triangle, matching the circle’s diameter with the hypotenuse a + b (Problem 4.7). The altitude cannot exceed the radius; therefore,

图像

此外,只有当三角形的高也是半圆的半径时,即a = b时,两条边才相等。因此,这幅图将不等式及其相等条件包含在一个易于理解的图形中。(AM-GM不等式的另一种图解证明在习题4.33中展开。)

Furthermore, the two sides are equal only when the altitude of the triangle is also a radius of the semicircle—namely when a = b. The picture therefore contains the inequality and its equality condition in one easy-to-grasp object. (An alternative pictorial proof of the AM–GM inequality is developed in Problem 4.33.)

图像

问题 4.6 画一个三角形的外接圆

Problem 4.6 Circumscribing a circle around a triangle

这里有几个例子,显示了一个圆与一个三角形的外接关系。

Here are a few examples showing a circle circumscribed around a triangle.

图像

画图表明圆由三角形唯一确定。

Draw a picture to show that the circle is uniquely determined by the triangle.

问题 4.7 寻找正确的半圆

Problem 4.7 Finding the right semicircle

一个三角形唯一地确定了它的外接圆(问题 4.6)。然而,圆的直径可能不与三角形的一条边对齐。一个半圆能否始终外接于一个直角三角形,同时使圆的直径与斜边对齐?

A triangle uniquely determines its circumscribing circle (Problem 4.6). However, the circle’s diameter might not align with a side of the triangle. Can a semicircle always be circumscribed around a right triangle while aligning the circle’s diameter along the hypotenuse?

问题 4.8 三个数的几何平均数

Problem 4.8 Geometric mean of three numbers

对于三个非负数,AM-GM 不等式为

For three nonnegative numbers, the AM–GM inequality is

图像

为什么与它的二数不等式相比,这个不等式不太可能有几何证明?(如果你找到了证明,请告诉我。)

Why is this inequality, in contrast to its two-number cousin, unlikely to have a geometric proof? (If you find a proof, let me know.)

4.2.3 应用

4.2.3 Applications

算术平均数和几何平均数在数学中有着广泛的应用。第一个应用是一个通常用导数求解的问题:将一段固定长度的篱笆折成一个矩形,围住最大的花园。

Arithmetic and geometric means have wide mathematical application. The first application is a problem more often solved with derivatives: Fold a fixed length of fence into a rectangle enclosing the largest garden.

图像 什么形状的矩形面积最大?

What shape of rectangle maximizes the area?

这个问题涉及两个量:一个固定的周长和一个需要最大化的面积。如果周长与算术平均值相关,面积与几何平均值相关,那么AM-GM不等式可能有助于最大化面积。周长P = 2(a + b)是算术平均值的四倍,面积A = ab是几何平均值的平方。因此,根据AM-GM不等式,

The problem involves two quantities: a perimeter that is fixed and an area to maximize. If the perimeter is related to the arithmetic mean and the area to the geometric mean, then the AM–GM inequality might help maximize the area. The perimeter P = 2(a + b) is four times the arithmetic mean, and the area A = ab is the square of the geometric mean. Therefore, from the AM–GM inequality,

图像

当 a = b 时,等式成立。左侧边由栅栏的长度决定。因此,右侧边随 a 和 b 的变化而变化,当 a = b 时,最大值为 P/4。面积最大的矩形是正方形。

with equality when a = b. The left side is fixed by the amount of fence. Thus the right side, which varies depending on a and b, has a maximum of P/4 when a = b. The maximal-area rectangle is a square.

图像

问题 4.9 直接图解证明

Problem 4.9 Direct pictorial proof

最大矩形花园的AM-GM推理是间接图形推理。它是建立在AM-GM不等式图形证明基础上的符号推理。你能画一幅图直接表明正方形是最优形状吗?

The AM–GM reasoning for the maximal rectangular garden is indirect pictorial reasoning. It is symbolic reasoning built upon the pictorial proof for the AM–GM inequality. Can you draw a picture to show directly that the square is the optimal shape?

问题 4.10 三部分产品

Problem 4.10 Three-part product

不使用微积分,求出 f(x) = x 2 (1 − 2x) 在 x ≥ 0 时的最大值。画出 f(x) 的简图来验证你的答案。

Find the maximum value of f(x) = x2(1 − 2x) for x ≥ 0, without using calculus. Sketch f(x) to confirm your answer.

问题 4.11 不受限制的最大面积

Problem 4.11 Unrestricted maximal area

如果花园不必是矩形,那么最大面积的形状是什么?

If the garden need not be rectangular, what is the maximal-area shape?

问题 4.12 体积最大化

Problem 4.12 Volume maximization

按照以下步骤制作一个顶部开口的盒子:从一个单位正方形开始,剪出四个相同的角,然后将盖子折叠起来。盒子的体积为 V = x(1 − 2x) 2,其中 x 是每个角的边长。选择哪个 x 可以使盒子的体积最大?

Build an open-topped box as follows: Start with a unit square, cut out four identical corners, and fold in the flaps. The box has volume V = x(1 − 2x)2, where x is the side length of a corner cutout. What choice of x maximizes the volume of the box?

图像

以下是基于矩形花园分析的合理分析。设 a = x,b = 1 − 2x,c = 1 − 2x。则 abc 为体积 V,V 1/3 =图像为几何平均值(问题 4.8)。由于几何平均值永远不会超过算术平均值,并且当 a = b = c 时两个平均值相等,因此当 x = 1 − 2x 时体积最大。因此,选择 x = 1/3 应该可以使盒子的体积最大化。

Here is a plausible analysis modeled on the analysis of the rectangular garden. Set a = x, b = 1 − 2x, and c = 1 − 2x. Then abc is the volume V, and V1/3 = is the geometric mean (Problem 4.8). Because the geometric mean never exceeds the arithmetic mean and because the two means are equal when a = b = c, the maximum volume is attained when x = 1 − 2x. Therefore, choosing x = 1/3 should maximize the volume of the box.

现在通过绘制 V(x) 或设置 dV/dx = 0 来表明这个选择是错误的;解释前面的推理有什么错误;并给出正确的版本。

Now show that this choice is wrong by graphing V(x) or setting dV/dx = 0; explain what is wrong with the preceding reasoning; and make a correct version.

问题 4.13 三角最小值

Problem 4.13 Trigonometric minimum

找到最小值

Find the minimum value of

图像

在区域 x ∈ (0, π) 内。

in the region x ∈ (0, π).

问题 4.14 三角最大值

Problem 4.14 Trigonometric maximum

在区域 t ∈ [0, π/2] 内,最大化 sin 2t,或者等效地,2 sin t cost t。

In the region t ∈ [0, π/2], maximize sin 2t or, equivalently, 2 sin t cos t.

算术平均数和几何平均数的第二个应用是计算π的现代方法,这种方法速度惊人地快[ 5,6 ]。古代计算π的方法包括计算多边正多边形的周长,精度可达小数点后几位。

The second application of arithmetic and geometric means is a modern, amazingly rapid method for computing π [5, 6]. Ancient methods for computing π included calculating the perimeter of many-sided regular polygons and provided a few decimal places of accuracy.

最近的计算使用了莱布尼茨的反正切级数

Recent computations have used Leibniz’s arctangent series

图像

假设你想计算π的10 ^9位数字,或许是为了测试一台新型超级计算机的硬件,或者研究π的数值是否随机(这是卡尔·萨根小说《接触》 [ 40 ]的主题)。在莱布尼茨级数中,设x = 1,结果为π/4,但该级数收敛速度极慢。得到10^ 9位数字大约需要图像几个项——比宇宙中的原子总数还要多得多。

Imagine that you want to compute π to 109 digits, perhaps to test the hardware of a new supercomputer or to study whether the digits of π are random (a theme in Carl Sagan’s novel Contact [40]). Setting x = 1 in the Leibniz series produces π/4, but the series converges extremely slowly. Obtaining 109 digits requires roughly terms—far more terms than atoms in the universe.

幸运的是,约翰·梅钦 (John Machin) (1686–1751) 提出了一个令人惊讶的三角恒等式

Fortunately, a surprising trigonometric identity due to John Machin (1686–1751)

图像

通过减少 x 来加速收敛:

accelerates the convergence by reducing x:

图像

即使加速,10 9位的精度也需要计算大约 10 9 个项。

Even with the speedup, 109-digit accuracy requires calculating roughly 109 terms.

相比之下,现代的布伦特-萨拉明算法 [ 3 , 41 ] 依赖于算术平均值和几何平均值,能够非常快速地收敛到π。该算法与计算椭圆周长(问题 4.15)以及计算互感的惊人精确方法密切相关 [ 23 ]。该算法首先从 a 0 = 1 和 g 0 = 1开始生成多个序列图像,然后计算连续的算术平均值 a n、几何平均值 g n及其平方差 d n

In contrast, the modern Brent–Salamin algorithm [3, 41], which relies on arithmetic and geometric means, converges to π extremely rapidly. The algorithm is closely related to amazingly accurate methods for calculating the perimeter of an ellipse (Problem 4.15) and also for calculating mutual inductance [23]. The algorithm generates several sequences by starting with a0 = 1 and g0 = 1/ it then computes successive arithmetic means an, geometric means gn, and their squared differences dn.

图像

a 和 g 序列迅速收敛到一个数 M(a 0 , g 0 ),该数称为 a 0和 g 0的算术几何平均值。然后,M(a 0 , g 0 ) 和差序列 d 确定 π。

The a and g sequences rapidly converge to a number M(a0, g0) called the arithmetic–geometric mean of a0 and g0. Then M(a0, g0) and the difference sequence d determine π.

图像

d 序列以二次函数形式趋近于零;换句话说,图像问题 4.16)。因此,在计算 π 时,每次迭代都会使精度位数翻倍。计算十亿位的 π 只需要大约 30 次迭代——远少于图像使用 x = 1 的反正切级数的项数,甚至少于使用 Machin 加速比的 10 ^9项。

The d sequence approaches zero quadratically; in other words, (Problem 4.16). Therefore, each iteration in this computation of π doubles the digits of accuracy. A billion-digit calculation of π requires only about 30 iterations—far fewer than the terms using the arctangent series with x = 1 or even than the 109 terms using Machin’s speedup.

问题 4.15 椭圆的周长

Problem 4.15 Perimeter of an ellipse

要计算长半轴为 a 0、短半轴为 g 0的椭圆的周长,需要计算 a、g 和 d 序列,以及 a 和 g 序列的公共极限 M(a 0 , g 0 ),就像计算 π 一样。然后,周长 P 可以通过以下公式计算:

To compute the perimeter of an ellipse with semimajor axis a0 and semiminor axis g0, compute the a, g, and d sequences and the common limit M(a0, g0) of the a and g sequences, as for the computation of π. Then the perimeter P can be computed with the following formula:

图像

其中 A 和 B 是需要确定的常数。使用简单情况法(第二章)确定它们的值。(请参阅 [ 3 ] 检查你的值并查看完整公式的证明。)

where A and B are constants for you to determine. Use the method of easy cases (Chapter 2) to determine their values. (See [3] to check your values and for a proof of the completed formula.)

问题 4.16 二次收敛

Problem 4.16 Quadratic convergence

从 a 0 = 1 和 g 0 = 1/ 图像(或任何其他正对)开始,然后按照 AM-GM 序列进行几次迭代

Start with a0 = 1 and g0 = 1/ (or any other positive pair) and follow several iterations of the AM–GM sequence

图像

然后生成图像图像检查图像(二次收敛)。

Then generate and to check that (quadratic convergence).

问题 4.17 收敛速度

Problem 4.17 Rapidity of convergence

选择一个正的 x 0;然后通过迭代生成一个序列

Pick a positive x0; then generate a sequence by the iteration

图像

序列收敛到什么?收敛速度有多快?如果 x 0 < 0 会怎样?

To what and how rapidly does the sequence converge? What if x0 < 0?

4.3 近似对数

4.3 Approximating the logarithm

函数通常用泰勒级数来近似

A function is often approximated by its Taylor series

图像

这看起来像一串不直观的符号。幸运的是,图片通常可以解释函数近似中最重要的几个项。例如,单项近似 sin θ ≈ θ,用圆的弧代替三角形的高,将非线性摆微分方程转化为易于处理的线性方程(见第 3.5 节)。

which looks like an unintuitive sequence of symbols. Fortunately, pictures often explain the first and most important terms in a function approximation. For example, the one-term approximation sin θ ≈ θ, which replaces the altitude of the triangle by the arc of the circle, turns the nonlinear pendulum differential equation into a tractable, linear equation (Section 3.5).

图像

另一个关于图像值的泰勒级数说明来自对数函数的级数:

Another Taylor-series illustration of the value of pictures come from the series for the logarithm function:

图像

它的第一项 x 可以导出近似值 (1 + x) n ≈ e nx (其中 x 较小且 n 任意) 的精确表达式(见 5.3.4 节)。它的第二项 −x 2 /2 有助于评估该近似值的精度。前两项是最有用的,并且有图示说明。

Its first term, x, will lead to the wonderful approximation (1 + x)n ≈ enx for small x and arbitrary n (Section 5.3.4). Its second term, −x2/2, helps evaluate the accuracy of that approximation. These first two terms are the most useful terms—and they have pictorial explanations.

起始图像是积分表示

The starting picture is the integral representation

图像

图像

图像 阴影区域最简单的近似值是什么?

What is the simplest approximation for the shaded area?

初步近似地,阴影区域大致相当于外接矩形——这是集中计算的一个例子。该矩形的面积为 x:

As a first approximation, the shaded area is roughly the circumscribed rectangle—an example of lumping. The rectangle has area x:

图像

图像

此区域重现了泰勒级数的第一项。由于使用了外接矩形,因此略微高估了 ln(1 + x)。

This area reproduces the first term in the Taylor series. Because it uses a circumscribed rectangle, it slightly overestimates ln(1 + x).

面积也可以通过画一个内接矩形来近似计算。它的宽度仍然是 x,但高度不是 1,而是 1/(1 + x),近似等于 1 − x(问题 4.18)。因此,内接矩形的面积近似为 x(1 − x) = x − x 2。这个面积略微低估了 ln(1 + x)。

The area can also be approximated by drawing an inscribed rectangle. Its width is again x, but its height is not 1 but rather 1/(1 + x), which is approximately 1 − x (Problem 4.18). Thus the inscribed rectangle has the approximate area x(1 − x) = x − x2. This area slightly underestimates ln(1 + x).

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问题 4.18 倒数函数的近似图

Problem 4.18 Picture for approximating the reciprocal function

确认近似值

Confirm the approximation

图像

尝试 x = 0.1 或 x = 0.2。然后画图说明等效近似 (1 − x)(1 + x) ≈ 1。

by trying x = 0.1 or x = 0.2. Then draw a picture to illustrate the equivalent approximation (1 − x)(1 + x) ≈ 1.

现在,我们得到了 ln(1 + x) 的两个近似值。第一个近似值稍微简单一些,它来自于绘制外接矩形。第二个近似值来自于绘制内接矩形。这两个近似值都围绕着精确值波动。

We now have two approximations to ln(1 + x). The first and slightly simpler approximation came from drawing the circumscribed rectangle. The second approximation came from drawing the inscribed rectangle. Both dance around the exact value.

图像 如何将内接矩形和外接矩形近似结合起来以获得更好的近似?

How can the inscribed-and circumscribed-rectangle approximations be combined to make an improved approximation?

一个近似值高估了面积,另一个近似值低估了面积;它们的平均值应该比任何一个近似值都要好。平均值是一个梯形,面积为

One approximation overestimates the area, and the other underestimates the area; their average ought to improve on either approximation. The average is a trapezoid with area

图像

图像

此区域重现了完整泰勒级数的前两项

This area reproduces the first two terms of the full Taylor series

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问题 4.19 三次项

Problem 4.19 Cubic term

通过估计梯形和真实面积之间的差异来估计泰勒级数中的三次项。

Estimate the cubic term in the Taylor series by estimating the difference between the trapezoid and the true area.

对于这些对数近似,最难的问题是 ln 2。

For these logarithm approximations, the hardest problem is ln 2.

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两个近似值都与真实值(约为 0.693)相差很大。即使是 ln(2) 的中等精度,也需要许多泰勒级数项,远远超出了图片所能解释的范围(问题 4.20)。问题在于 ln(1 + x) 中的 x 等于 1,因此泰勒级数每一项中的 x n因子并不会缩小高 n 项。

Both approximations differ significantly from the true value (roughly 0.693). Even moderate accuracy for ln 2 requires many terms of the Taylor series, far beyond what pictures explain (Problem 4.20). The problem is that x in ln(1 + x) is 1, so the xn factor in each term of the Taylor series does not shrink the high-n terms.

使用莱布尼茨反正切级数计算 π 时也会出现同样的问题(第 4.2.3 节

The same problem happens when computing π using Leibniz’s arctangent series (Section 4.2.3)

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假设 x = 1,π/4 的直接近似值需要很多项才能达到中等精度。幸运的是,三角恒等式 arctan 1 = 4 arctan 1/5 − arctan 1/239 将最大 x 降低到 1/5,从而加快了收敛速度。

By using x = 1, the direct approximation of π/4 requires many terms to attain even moderate accuracy. Fortunately, the trigonometric identity arctan 1 = 4 arctan 1/5 − arctan 1/239 lowers the largest x to 1/5 and thereby speeds the convergence.

图像 是否有类似物可以帮助估计ln 2?

Is there an analogous that helps estimate ln 2?

因为 2 也是 (4/3)/(2/3),所以 ln 2 的类似重写为

Because 2 is also (4/3)/(2/3), an analogous rewriting of ln 2 is

图像

每个分数的形式都是 1 + x,其中 x = ±1/3。由于 x 较小,对数级数的一项可能就能提供合理的精度。因此,我们用 ln(1 + x) ≈ x 来近似这两个对数:

Each fraction has the form 1 + x with x = ±1/3. Because x is small, one term of the logarithm series might provide reasonable accuracy. Let’s therefore use ln(1 + x) ≈ x to approximate the two logarithms:

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这个估计准确度在5%以内!

This estimate is accurate to within 5%!

重写技巧有助于计算π(通过重写arctan x级数)和估算ln(1 + x)(通过重写x本身)。因此,这个想法变成了一种方法——一个我使用过两次的技巧(这个定义经常被认为是波利亚的)。

The rewriting trick has helped to compute π (by rewriting the arctan x series) and to estimate ln(1 + x) (by rewriting x itself). This idea therefore becomes a method—a trick that I use twice (this definition is often attributed to Polya).

问题 4.20 有多少项?

Problem 4.20 How many terms?

对数的完整泰勒级数是

The full Taylor series for the logarithm is

图像

如果在此系列中设置 x = 1,则需要多少项才能将 ln 2 估计到 5% 以内?

If you set x = 1 in this series, how many terms are required to estimate ln 2 to within 5%?

问题 4.21 第二次重写

Problem 4.21 Second rewriting

重复重写法,将 4/3 和 2/3 重写;然后仅使用对数级数的一项来估计 ln2。修订后的估计值有多准确?

Repeat the rewriting method by rewriting 4/3 and 2/3; then estimate ln 2 using only one term of the logarithm series. How accurate is the revised estimate?

问题 4.22 泰勒级数的两个项

Problem 4.22 Two terms of the Taylor series

将 ln 2 重写为 ln(4/3) − ln(2/3),使用两项近似值 ln(1+x) ≈ x−x 2 /2 来估计 ln 2 。将该近似值与一项估计值 2/3 进行比较。(问题 4.24探讨了图形解释。)

After rewriting ln 2 as ln(4/3) − ln(2/3), use the two-term approximation that ln(1+x) ≈ x−x2/2 to estimate ln 2. Compare the approximation to the one-term estimate, namely 2/3. (Problem 4.24 investigates a pictorial explanation.)

问题 4.23 对数的有理函数近似

Problem 4.23 Rational-function approximation for the logarithm

替换 ln 2 = ln(4/3) − ln(2/3) 具有一般形式

The replacement ln 2 = ln(4/3) − ln(2/3) has the general form

图像

其中 y = x/(2 + x)。

where y = x/(2 + x).

利用 y 的表达式和单项级数 ln(1+x) ≈ x,将 ln(1+x) 表示为 x 的有理函数(即 x 的多项式比)。其泰勒级数的前几项是什么?

Use the expression for y and the one-term series ln(1+x) ≈ x to express ln(1+x) as a rational function of x (as a ratio of polynomials in x). What are the first few terms of its Taylor series?

将这些项与 ln(1 + x) 泰勒级数的前几项进行比较,从而解释为什么有理函数近似比两项级数 ln(1 + x) ≈ x − x 2 /2 更准确。

Compare those terms to the first few terms of the ln(1 + x) Taylor series, and thereby explain why the rational-function approximation is more accurate than even the two-term series ln(1 + x) ≈ x − x2/2.

问题 4.24 重写的图解

Problem 4.24 Pictorial interpretation of the rewriting

a. 使用 ln(1 + x) 的积分表示来解释为什么阴影面积是 ln 2。

a. Use the integral representation of ln(1 + x) to explain why the shaded area is ln 2.

b. 勾勒出代表的区域

b. Outline the region that represents

图像

当对每个对数使用外接矩形近似时。

when using the circumscribed-rectangle approximation for each logarithm.

图像

c. 使用梯形近似ln(1+x) = x−x 2 /2,勾勒出同一区域。用图表示,虽然该区域形状不同,但其面积与你在b项中绘制的区域相同。

c. Outline the same region when using the trapezoid approximation ln(1+x) = x−x2/2. Show pictorially that this region, although a different shape, has the same area as the region that you drew in item b.

4.4 平分三角形

4.4 Bisecting a triangle

对于几何问题来说,图形解决方案尤其适用:

Pictorial solutions are especially likely for a geometric problem:

图像 将等边三角形一分为二成两个面积相等的区域的最短路径是什么?

What is the shortest path that bisects an equilateral triangle into two regions of equal area?

可能的二等分路径构成了一个不可数的无限集合。为了应对复杂性,可以尝试一些简单的情况(第二章)——画几个等边三角形,然后用简单的路径将它们二等分。模式、想法,甚至解决方案都可能浮现出来。

The possible bisecting paths form an uncountably infinite set. To manage the complexity, try easy cases (Chapter 2)—draw a few equilateral triangles and bisect them with easy paths. Patterns, ideas, or even a solution might emerge.

图像 有哪些简单的路径?

What are a few easy paths?

最简单的平分路径是一条垂直线段,将三角形分成两个底边各为 1/2 的直角三角形。这条路径就是三角形的高,长度为

The simplest bisecting path is a vertical segment that splits the triangle into two right triangles each with base 1/2. This path is the triangle’s altitude, and it has length

图像

图像

另一条直线路径将三角形分成一个梯形和一个小三角形。

An alternative straight path splits the triangle into a trapezoid and a small triangle.

图像 较小三角形的形状是什么?路径有多长?

What is the shape of the smaller triangle, and how long is the path?

图像

这个三角形与原来的三角形相似,所以它也是等边三角形。此外,它的面积是原来的一半,所以它的三个其中一条边是平分路径,其长度图像比原始三角形的边长小 。因此,这条路径的长度为 1/ 图像≈ 0.707——比长度为 的垂直路径有了显著的改进图像

The triangle is similar to the original triangle, so it too is equilateral. Furthermore, it has one-half of the area of the original triangle, so its three sides, one of which is the bisecting path, are a factor of smaller than the sides of the original triangle. Thus this path has length 1/ ≈ 0.707—a substantial improvement on the vertical path with length .

问题 4.25 所有单段路径

Problem 4.25 All one-segment paths

等边三角形有无数条一条线段平分路径。图中显示了其中几条。哪条一条线段路径最短?

An equilateral triangle has infinitely many one-segment bisecting paths. A few of them are shown in the figure. Which one-segment path is the shortest?

图像

现在让我们研究一下简单的两段路径。一条可能的路径包含一个菱形,但不包括两个小三角形。这两个小三角形占据了整个区域的一半。因此,每个小三角形占据了整个区域的四分之一,边长为 1/2。由于平分路径包含其中两条边,因此其长度为 1。不幸的是,这条路径比我们之前提到的两个一段路径候选路径要长,后者的长度分别为 1/图像图像

Now let’s investigate easy two-segment paths. One possible path encloses a diamond and excludes two small triangles. The two small triangles occupy one-half of the entire area. Each small triangle therefore occupies one-fourth of the entire area and has side length 1/2. Because the bisecting path contains two of these sides, it has length 1. This path is, unfortunately, longer than our two one-segment candidates, whose lengths are 1/ and .

图像

因此,一个合理的猜想是,最短路径具有最少的段数。这个猜想值得验证(问题 4.26)。

Therefore, a reasonable conjecture is that the shortest path has the fewest segments. This conjecture deserves to be tested (Problem 4.26).

问题 4.26 所有两段路径

Problem 4.26 All two-segment paths

画一个图来表示两段路径的多样性。找出最短路径,并说明其长度

Draw a figure showing the variety of two-segment paths. Find the shortest path, showing that it has length

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问题 4.27 封闭路径二等分

Problem 4.27 Bisecting with closed paths

平分路径不必以三角形的边为起点或终点。以下是两个示例:

The bisecting path need not begin or end at an edge of the triangle. Two examples are illustrated here:

图像

你认为闭合平分路径会比最短单段路径更长还是更短?请给出你猜想的几何依据,并通过计算两条示例闭合路径的长度来验证猜想。

Do you expect closed bisecting paths to be longer or shorter than the shortest one-segment path? Give a geometric reason for your conjecture, and check the conjecture by finding the lengths of the two illustrative closed paths.

图像 使用较少的段是否会产生较短的路径?

Does using fewer segments produce shorter paths?

最短的单段路径长度约为 0.707;但最短的两段路径长度约为 0.681。长度的缩短意味着可以尝试极端路径:包含无限多条线段。换句话说,尝试曲线路径。最简单的曲线路径可能是圆形或圆形的一部分。

The shortest one-segment path has an approximate length of 0.707; but the shortest two-segment path has an approximate length of 0.681. The length decrease suggests trying extreme paths: paths with an infinite number of segments. In other words, try curved paths. The easiest curved path is probably a circle or a piece of a circle.

图像 可以将三角形一分为二的最短圆或圆的一部分最有可能是什么?

What is a likely candidate for the shortest circle or piece of a circle that bisects the triangle?

无论路径是圆形还是圆形的一部分,都需要一个圆心。然而,将圆心置于三角形内部并使用完整的圆,会产生一条较长的平分路径(问题 4.27)。唯一可能的另一个圆心是三角形的一个顶点,因此可以想象一条以该顶点为中心的平分弧。

Whether the path is a circle or piece of a circle, it needs a center. However, putting the center inside the triangle and using a full circle produces a long bisecting path (Problem 4.27). The only other plausible center is a vertex of the triangle, so imagine a bisecting arc centered on one vertex.

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图像 这个弧有多长?

How long is this arc?

圆弧所对的圆的六分之一(60°),因此其长度为 l = πr/3,其中 r 是圆的半径。要计算半径,需要利用圆弧必须平分三角形的条件。因此,圆弧围住了三角形面积的一半。r 的条件是 πr 2 =图像

The arc subtends one-sixth (60°) of the full circle, so its length is l = πr/3, where r is radius of the full circle. To find the radius, use the requirement that the arc must bisect the triangle. Therefore, the arc encloses one-half of the triangle’s area. The condition on r is that πr2 =

图像

因此,半径为图像;圆弧长度为πr/3,约为0.673。这条曲线比最短的两段路径更短。它可能是可能的最短路径。

The radius is therefore ; the length of the arc is πr/3, which is approximately 0.673. This curved path is shorter than the shortest two-segment path. It might be the shortest possible path.

为了验证这个猜想,我们利用对称性。因为等边三角形是六边形的六分之一,所以复制被平分的等边三角形可以构建一个六边形。下图是由被水平线平分的三角形构建的六边形:

To test this conjecture, we use symmetry. Because an equilateral triangle is one-sixth of a hexagon, build a hexagon by replicating the bisected equilateral triangle. Here is the hexagon built from the triangle bisected by a horizontal line:

图像

六条平分路径形成一个内部六边形,其面积是大六边形面积的一半。

The six bisecting paths form an internal hexagon whose area is one-half of the area of the large hexagon.

图像 复制圆弧平分的三角形会发生什么?

What happens when replicating the triangle bisected by the circular arc?

当复制该三角形时,六个副本构成一个圆,其面积等于六边形面积的一半。对于给定的面积,圆的周长最短(等周定理[ 30 ]和问题4.11);因此,圆的六分之一是最短的平分路径。

When that triangle is replicated, its six copies make a circle with area equal to one-half of the area of the hexagon. For a fixed area, a circle has the shortest perimeter (the isoperimetric theorem [30] and Problem 4.11); therefore, one-sixth of the circle is the shortest bisecting path.

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问题 4.28 复制垂直二等分

Problem 4.28 Replicating the vertical bisection

如果复制一条垂直线平分的三角形,并且只旋转它,就会得到一个破碎的封闭区域,而不是凸多边形。如何复制这个三角形,使六条平分路径形成一个正多边形?

The triangle bisected by a vertical line, if replicated and only rotated, produces a fragmented enclosed region rather than a convex polygon. How can the triangle be replicated so that the six bisecting paths form a regular polygon?

问题 4.29 平分立方体

Problem 4.29 Bisecting the cube

在将立方体一分为二为两个相等体积的所有表面中,哪个表面的面积最小?

Of all surfaces that bisect a cube into two equal volumes, which surface has the smallest area?

4.5 求和级数

4.5 Summing series

最后一个例子,说明图片可以解释什么,回到阶乘函数。我们对 n! 的第一次近似,首先从它的积分表示开始,然后使用集中法(第 3.2.3 节)。

For the final example of what pictures can explain, return to the factorial function. Our first approximation to n! began with its integral representation and then used lumping (Section 3.2.3).

通过将曲线替换为易于计算面积的矩形,这种集中方法本身就是一种图形分析。n! 的第二幅图从求和表示开始

Lumping, by replacing a curve with a rectangle whose area is easily computed, is already a pictorial analysis. A second picture for n! begins with the summation representation

图像

图像

该总和等于外接矩形的总面积。

This sum equals the combined area of the circumscribing rectangles.

问题 4.30 绘制平滑曲线

Problem 4.30 Drawing the smooth curve

设置矩形的高度需要绘制 ln k 曲线——该曲线可能与每个矩形的顶边沿任意位置相交。在上图和本节的分析中,曲线与边的右端点相交。阅读本节内容后,请针对另外两种情况重新进行分析:

Setting the height of the rectangles requires drawing the ln k curve—which could intersect the top edge of each rectangle anywhere along the edge. In the preceding figure and the analysis of this section, the curve intersects at the right endpoint of the edge. After reading the section, redo the analysis for two other cases:

a. 曲线与边的左端点相交。

a. The curve intersects at the left endpoint of the edge.

b. 曲线与边的中点相交。

b. The curve intersects at the midpoint of the edge.

该合并面积大约等于 ln k 曲线下的面积,因此

That combined area is approximately the area under the ln k curve, so

图像

图像

此 ln n! 近似值中的每个项都会为 n! 贡献一个因子:

Each term in this ln n! approximation contributes one factor to n!:

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每个因子在斯特林近似中都有一个对应的因子(见第 3.2.3 节)。按重要性降序排列,斯特林近似中的因子如下:

Each factor has a counterpart in a factor from Stirling’s approximation (Section 3.2.3). In descending order of importance, the factors in Stirling’s approximation are

图像

积分近似值重现了两个最重要的因子,并且几乎重现了第四个因子:e 和图像仅相差 8%。唯一无法解释的因子是图像

The integral approximation reproduces the two most important factors and almost reproduces the fourth factor: e and differ by only 8%. The only unexplained factor is

图像 这个因素从何图像而来?

From where does the factor come?

图像因子肯定来自ln k曲线上方的片段。它们几乎是三角形,如果是三角形的话,添加起来会更容易。因此,使用直线段重新绘制ln k曲线(另一种集中法)。

The factor must come from the fragments above the ln k curve. They are almost triangles and would be easier to add if they were triangles. Therefore, redraw the ln k curve using straight-line segments (another use of lumping).

图像

如果生成的三角形是矩形,那么它们会更容易相加。因此,我们将每个三角形加倍,使其成为一个矩形。

The resulting triangles would be easier to add if they were rectangles. Therefore, let’s double each triangle to make it a rectangle.

图像

图像 这些矩形块的总和是多少?

What is the sum of these rectangular pieces?

要将这些碎片相加,请将右手沿着 k = n 的垂直线放置。用左手将这些碎片向右推,直到它们碰到你的右手。然后,这些碎片堆叠起来,形成一个 ln n 矩形。由于每个碎片的长度都是相应三角形凸起部分的两倍,因此三角形凸起部分的总和等于 (ln n)/2。这种三角形修正改进了积分近似值。由此得到的 ln n! 近似值现在多了一项:

To sum these pieces, lay your right hand along the k = n vertical line. With your left hand, shove the pieces to the right until they hit your right hand. The pieces then stack to form the ln n rectangle. Because each piece is double the corresponding triangular protrusion, the triangular protrusions sum to (ln n)/2. This triangle correction improves the integral approximation. The resulting approximation for ln n! now has one more term:

图像

图像

经过指数运算得到 n!,修正贡献了一个因子图像

Upon exponentiating to get n!, the correction contributes a factor of

图像

与斯特林近似值相比,唯一剩下的区别是应该是的因子 e 图像,误差仅为 8%——只需进行一次积分并绘制几张图即可。

Compared to Stirling’s approximation, the only remaining difference is the factor of e that should be , an error of only 8%—all from doing one integral and drawing a few pictures.

问题 4.31 低估还是高估?

Problem 4.31 Underestimate or overestimate?

用三角形修正的积分近似法是低估了还是高估了n!?先用图形推理,然后用数值方法检验结论。

Does the integral approximation with the triangle correction underestimate or overestimate n!? Use pictorial reasoning; then check the conclusion numerically.

问题 4.32 下一个修正

Problem 4.32 Next correction

三角形校正是一系列无限校正中的第一个。这些校正包含与 n −2、n −3 ……成比例的项,仅凭图片很难推导出来。但 n −1校正可以用图片推导出来。

The triangle correction is the first of an infinite series of corrections. The corrections include terms proportional to n−2, n−3, . . ., and they are difficult to derive using only pictures. But the n−1 correction can be derived with pictures.

a. 绘制表示用分段线性曲线(由直线段组成的曲线)替换平滑的 ln k 曲线所产生的误差的区域。

a. Draw the regions showing the error made by replacing the smooth ln k curve with a piecewise-linear curve (a curve made of straight segments).

b. 每个区域上方都有一条近似抛物线的曲线,其面积由阿基米德公式给出(问题 4.34

b. Each region is bounded above by a curve that is almost a parabola, whose area is given by Archimedes’ formula (Problem 4.34)

图像

使用该属性来近似计算每个区域的面积。

Use that property to approximate the area of each region.

c. 证明当计算 ln n! = 时,图像这些区域之和约为 (1 − n −1 )/12。

c. Show that when evaluating ln n! = these regions sum to approximately (1 − n−1)/12.

d. 在 n! 近似中,改进后的常数项(以前为 e)是多少?它与ln n! 近似中的图像n −1项对 n! 近似的贡献有多大?

d. What is the resulting, improved constant term (formerly e) in the approximation to n! and how close is it to What factor does the n−1 term in the ln n! approximation contribute to the n! approximation?

这些修正和后续修正是在第 6.3.2 节中使用类比技术得出的。

These and subsequent corrections are derived in Section 6.3.2 using the technique of analogy.

4.6 总结和进一步的问题

4.6 Summary and further problems

数千万年来,进化不断完善着我们的感知能力。小孩子比成年人更能准确、更迅速地识别各种模式。最大的超级计算机。因此,图像推理能够利用大脑强大的计算能力。它帮助我们一目了然地理解和洞察宏大概念,从而让我们更加聪明。

For tens of millions of years, evolution has refined our perceptual abilities. A small child recognizes patterns more reliably and quickly than does the largest supercomputer. Pictorial reasoning, therefore, taps the mind’s vast computational power. It makes us more intelligent by helping us understand and see large ideas at a glance.

想要获取大量有趣的图片证明,请参阅 Nelsen 的著作 [ 31 , 32 ]。以下是一些进一步发展图像推理的问题。

For extensive and enjoyable collections of picture proofs, see the works of Nelsen [31, 32]. Here are further problems to develop pictorial reasoning.

问题 4.33 AM-GM 不等式的另一幅图

Problem 4.33 Another picture for the AM–GM inequality

绘制 y = ln x 来表明 a 和 b 的算术平均值始终大于或等于它们的几何平均值,并且当 a = b 时相等。

Sketch y = ln x to show that the arithmetic mean of a and b is always greater than or equal to their geometric mean, with equality when a = b.

问题 4.34 阿基米德抛物线面积公式

Problem 4.34 Archimedes’ formula for the area of a parabola

阿基米德证明(早在微积分出现之前!),闭抛物线包含其外接矩形的三分之二。用积分法证明此结果。

Archimedes showed (long before calculus!) that the closed parabola encloses two-thirds of its circumscribing rectangle. Prove this result by integration.

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证明闭抛物线也能包含边垂直的外接平行四边形的三分之二。这些图示方法在近似函数时很有用(例如,在问题 4.32中)。

Show that the closed parabola also encloses two-thirds of the circumscribing parallelogram with vertical sides. These pictorial recipes are useful when approximating functions (for example, in Problem 4.32).

问题 4.35 圆的面积的古代图画

Problem 4.35 Ancient picture for the area of a circle

古希腊人知道半径为r的圆的周长是2πr。然后他们用下图证明它的面积是πr² 你能重现这个论证吗?

The ancient Greeks knew that the circumference of a circle with radius r was 2πr. They then used the following picture to show that its area is πr2. Can you reconstruct the argument?

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问题 4.36 球体的体积

Problem 4.36 Volume of a sphere

扩展问题 4.35的论证,求半径为 r 的球体的体积,已知其表面积为 4πr² 用草图说明该论证。

Extend the argument of Problem 4.35 to find the volume of a sphere of radius r, given that its surface area is 4πr2. Illustrate the argument with a sketch.

问题 4.37 一个著名的和

Problem 4.37 A famous sum

使用图形推理来近似著名的巴塞尔和图像

Use pictorial reasoning to approximate the famous Basel sum

问题 4.38 牛顿-拉夫森法

Problem 4.38 Newton–Raphson method

一般来说,求解 f(t) = 0 需要近似值。一种方法是从猜测 t 0开始,然后使用牛顿-拉夫森法迭代改进它

In general, solving f(t) = 0 requires approximations. One method is to start with a guess t0 and to improve it iteratively using the Newton–Raphson method

图像

其中 f (t n ) 是在 t = t n时刻求导的 df/dt 。画图证明此公式的合理性;然后使用此公式进行估算图像(然后尝试解答问题 4.17)。

where f(tn) is the derivative df/dt evaluated at t = tn. Draw a picture to justify this recipe; then use the recipe to estimate (Then try Problem 4.17.)

5

5

取出大部件

Taking out the big part

在几乎所有定量问题中,只要遵循“先做要事”这一谚语,分析就会变得简单。首先近似并理解最重要的影响因素——最重要的部分——然后完善你的分析和理解。这种逐次逼近或“去掉最重要的部分”的过程可以生成有意义、令人难忘且可用的表达式。以下示例介绍了低熵表达式的相关概念(第 5.2 节),并分析了心算乘法(第 5.1 节)、指数运算(第 5.3 节)、二次方程(第 5.4 节)以及一个困难的三角积分(第 5.5 节)。

In almost every quantitative problem, the analysis simplifies when you follow the proverbial advice of doing first things first. First approximate and understand the most important effect—the big part—then refine your analysis and understanding. This procedure of successive approximation or “taking out the big part” generates meaningful, memorable, and usable expressions. The following examples introduce the related idea of low-entropy expressions (Section 5.2) and analyze mental multiplication (Section 5.1), exponentiation (Section 5.3), quadratic equations (Section 5.4), and a difficult trigonometric integral (Section 5.5).

5.1 使用“一”和“少”进行乘法

5.1 Multiplication using one and few

第一个例子是一种适合粗略估算的心算乘法。具体计算的是数据 CD-ROM 的存储容量。数据 CD-ROM 的格式和存储容量与音乐 CD 相同,其容量可以通过以下三个因素的乘积来估算:

The first illustration is a method of mental multiplication suited to rough, back-of-the-envelope estimates. The particular calculation is the storage capacity of a data CD-ROM. A data CD-ROM has the same format and storage capacity as a music CD, whose capacity can be estimated as the product of three factors:

图像

(在样本大小因素中,两个通道用于立体声。)

(In the sample-size factor, the two channels are for stereophonic sound.)

问题 5.1 采样率

Problem 5.1 Sample rate

查阅香农-奈奎斯特采样定理[ 22 ],并解释为什么采样率(测量声压的速率)大约为 40 kHz。

Look up the Shannon–Nyquist sampling theorem [22], and explain why the sample rate (the rate at which the sound pressure is measured) is roughly 40 kHz.

问题 5.2 每个样本的位数

Problem 5.2 Bits per sample

因为 2 16 ∼ 10 5,一个 16 位的采样(CD 格式选择)需要精确到大约 0.001% 的电子元件。为什么 CD 格式的设计者不选择更大的采样大小,比如 32 位(每通道)呢?

Because 216 ∼ 105, a 16-bit sample—as chosen for the CD format—requires electronics accurate to roughly 0.001%. Why didn’t the designers of the CD format choose a much larger sample size, say 32 bits (per channel)?

问题 5.3 检查单位

Problem 5.3 Checking units

检查估计中的所有单位是否都能除以所需比特单位。

Check that all the units in the estimate divide out—except for the desired units of bits.

粗略计算使用诸如播放时间之类的粗略估计,而忽略了诸如用于错误检测和纠正的比特数等重要因素。在这个估算以及许多其他估算中,使用精确到小数点后3位的乘法就显得有些过度了。近似分析需要近似的计算方法。

Back-of-the-envelope calculations use rough estimates such as the playing time and neglect important factors such as the bits devoted to error detection and correction. In this and many other estimates, multiplication with 3 decimal places of accuracy would be overkill. An approximate analysis needs an approximate method of calculation.

图像 数据容量在 2 倍以内是多少?

What is the data capacity to within a factor of 2?

单位(最大的部分!)是位(问题 5.3),三个数值因子贡献了 3600 × 4.4 × 10 4 × 32。要估算乘积,将其分成大部分和校正部分。

The units (the biggest part!) are bits (Problem 5.3), and the three numerical factors contribute 3600 × 4.4 × 104 × 32. To estimate the product, split it into a big part and a correction.

大数部分:粗略计算产品中最重要的因子通常来自 10 的幂,因此首先评估这个大数部分:3600 贡献 3 个 10 的幂,4.4 × 10 4贡献 4 个,32 贡献 1 个。10 的 8 个幂产生了10的因子。

The big part: The most important factor in a back-of-the-envelope product usually comes from the powers of 10, so evaluate this big part first: 3600 contributes three powers of 10, 4.4 × 104 contributes four, and 32 contributes one. The eight powers of 10 produce a factor of 108.

修正:去掉较大部分后,剩余部分是修正因子 3.6 × 4.4 × 3.2。此乘积也通过去掉较大部分简化。将每个因子四舍五入为三个选项中最接近的数字:1、few 或 10。我们发明的数字 few 介于 1 和 10 之间:它是 1 和 10 的几何平均值,因此 (few) 2 = 10,few ≈ 3。在乘积 3.6×4.4×3.2 中,每个因子四舍五入为 few,因此 3.6×4.4×3.2 ≈ (few) 3,大约等于 30。

The correction: After taking out the big part, the remaining part is a correction factor of 3.6 × 4.4 × 3.2. This product too is simplified by taking out its big part. Round each factor to the closest number among three choices: 1, few, or 10. The invented number few lies midway between 1 and 10: It is the geometric mean of 1 and 10, so (few)2 = 10 and few ≈ 3. In the product 3.6×4.4×3.2, each factor rounds to few, so 3.6×4.4×3.2 ≈ (few)3 or roughly 30.

单位、10 的幂和校正因子结合起来得出

The units, the powers of 10, and the correction factor combine to give

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该估计值与精确乘积(问题 5.4 )相差 2 倍以内,而精确乘积本身接近 5.6 × 10 9位的实际容量。

This estimate is within a factor of 2 of the exact product (Problem 5.4), which is itself close to the actual capacity of 5.6 × 109 bits.

问题 5.4 低估还是高估?

Problem 5.4 Underestimate or overestimate?

3 × 10 9是高估了还是低估了 3600 × 4.4 × 10 4 × 32 ?通过计算精确乘积来检验你的推理。

Does 3 × 109 overestimate or underestimate 3600 × 4.4 × 104 × 32? Check your reasoning by computing the exact product.

问题 5.5 更多练习

Problem 5.5 More practice

使用一次或几次乘法在心算中完成以下计算;然后比较近似值和实际乘积。

Use the one-or-few method of multiplication to perform the following calculations mentally; then compare the approximate and actual products.

a. 161 × 294 × 280 × 438。实际乘积约为 5.8 × 10 9

a. 161 × 294 × 280 × 438. The actual product is roughly 5.8 × 109.

b. 地球表面面积A = 4πR 2,其中半径R ∼ 6 × 10 6米。实际表面面积约为5.1 × 10 142

b. Earth’s surface area A = 4πR2, where the radius is R ∼ 6 × 106 m. The actual surface area is roughly 5.1 × 1014 m2.

5.2 分数变化和低熵表达式

5.2 Fractional changes and low-entropy expressions

使用“一或少数”法进行心算乘法很快。例如,3.15 × 7.21 很快就变成了“少数”× 10 1 ∼ 30,与精确乘积 22.7115 的误差在 50% 以内。为了得到更准确的估计,将 3.15 四舍五入为 3,将 7.21 四舍五入为 7。它们的乘积 21 的误差仅为 8%。为了进一步减少误差,可以将 3.15 × 7.21 分解成一个大数部分和一个加法修正部分。这种分解得到

Using the one-or-few method for mental multiplication is fast. For example, 3.15 × 7.21 quickly becomes few × 101 ∼ 30, which is within 50% of the exact product 22.7115. To get a more accurate estimate, round 3.15 to 3 and 7.21 to 7. Their product 21 is in error by only 8%. To reduce the error further, one could split 3.15 × 7.21 into a big part and an additive correction. This decomposition produces

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这种方法虽然合理,但直接应用去掉大数部分会产生混乱的修正,难以记忆和理解。不过,稍加修改,去掉大数部分就能得到清晰直观的修正。此外,改进后的修正方法还引入了两个重要的“街头斗殴”思想:分数变化(第 5.2.1 节)和低熵表达式(第 5.2.2 节)。改进后的修正方法将作为众多用途中的第一个,帮助我们估算高速公路限速节省的能量(第 5.2.3 节)。

The approach is sound, but the literal application of taking out the big part produces a messy correction that is hard to remember and understand. Slightly modified, however, taking out the big part provides a clean and intuitive correction. As gravy, developing the improved correction introduces two important street-fighting ideas: fractional changes (Section 5.2.1) and low-entropy expressions (Section 5.2.2). The improved correction will then, as a first of many uses, help us estimate the energy saved by highway speed limits (Section 5.2.3).

5.2.1 分数变化

5.2.1 Fractional changes

加法校正的卫生替代方法是将产品分成大部分和乘法校正:

The hygienic alternative to an additive correction is to split the product into a big part and a multiplicative correction:

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图像 你能找到校正因子的图片吗?

Can you find a picture for the correction factor?

校正因子是一个宽为 1 + 0.05、高为 1 + 0.03 的矩形的面积。该矩形包含 (1 + 0.05) × (1 + 0.03) 展开式中每一项对应的子矩形。它们的面积之和约为 1 + 0.05 + 0.03,表示比大部分增加了 8%。大部分为 21,其中 8% 为 1.68,因此 3.15 × 7.21 = 22.68,与精确乘积的误差在 0.14% 以内。

The correction factor is the area of a rectangle with width 1 + 0.05 and height 1 + 0.03. The rectangle contains one subrectangle for each term in the expansion of (1 + 0.05) × (1 + 0.03). Their combined area of roughly 1 + 0.05 + 0.03 represents an 8% fractional increase over the big part. The big part is 21, and 8% of it is 1.68, so 3.15 × 7.21 = 22.68, which is within 0.14% of the exact product.

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问题 5.6 分数误差图

Problem 5.6 Picture for the fractional error

大约 0.15% 的分数误差的图形解释是什么?

What is the pictorial explanation for the fractional error of roughly 0.15%?

问题 5.7 自己尝试一下

Problem 5.7 Try it yourself

将每个因数四舍五入到 10 的近端倍数,估算出 245×42,并将这个较大的部分与精确乘积进行比较。然后为校正因数画一个矩形,估算其面积,并校正这个较大的部分。

Estimate 245×42 by rounding each factor to a nearby multiple of 10, and compare this big part with the exact product. Then draw a rectangle for the correction factor, estimate its area, and correct the big part.

5.2.2 低熵表达式

5.2.2 Low-entropy expressions

3.15 × 7.21 的修正,作为绝对值或加性变化来说很复杂,但作为分数变化来说却很简单。这种对比具有普遍性。使用加性修正,双因子乘积变为

The correction to 3.15 × 7.21 was complicated as an absolute or additive change but simple as a fractional change. This contrast is general. Using the additive correction, a two-factor product becomes

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问题 5.8 矩形图片

Problem 5.8 Rectangle picture

画一个矩形代表扩展

Draw a rectangle representing the expansion

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当绝对变化 Δx 和 Δy 较小时(x 图像Δx 和 y 图像Δy),修正公式可简化为 xΔy + yΔx,但即便如此,它也很难记住,因为它包含许多看似合理但实际上不正确的替代项。例如,它可能包含 ΔxΔy、xΔx 或 yΔy 等项。合理替代方案的数量衡量了我们的直觉和现实之间的差距;差距越大,正确的结果就越难填补,我们也就越难记住正确的结果。

When the absolute changes Δx and Δy are small (x Δx and y Δy), the correction simplifies to xΔy + yΔx, but even so it is hard to remember because it has many plausible but incorrect alternatives. For example, it could plausibly contain terms such as ΔxΔy, xΔx, or yΔy. The extent of the plausible alternatives measures the gap between our intuition and reality; the larger the gap, the harder the correct result must work to fill it, and the harder we must work to remember the correct result.

此类差距是统计力学和信息论[20,21 ]的研究主题,它们将差距定义为合理替代方案数量的对数,并将对数值称为熵。对数并没有改变一个本质点:表达式在合理替代方案的数量上有所不同,而高熵表达式[ 28 ](具有许多合理替代方案的表达式)难以记忆和理解。

Such gaps are the subject of statistical mechanics and information theory [20, 21], which define the gap as the logarithm of the number of plausible alternatives and call the logarithmic quantity the entropy. The logarithm does not alter the essential point that expressions differ in the number of plausible alternatives and that high-entropy expressions [28]—ones with many plausible alternatives—are hard to remember and understand.

相比之下,低熵表达式几乎不允许出现任何合理的替代方案,并会引发“是的!怎么可能不是这样呢?!”许多数学和科学进步都在于找到将高熵表达式转化为易于理解的低熵表达式的思维方式。

In contrast, a low-entropy expression allows few plausible alternatives, and elicits, “Yes! How could it be otherwise?!” Much mathematical and scientific progress consists of finding ways of thinking that turn high-entropy expressions into easy-to-understand, low-entropy expressions.

图像 对乘积xy进行校正的低熵表达式是什么

What is a low-entropy expression for the correction to the product xy?

由于乘法校正是无量纲的,因此其熵值自然低于加法校正:合理的无量纲表达式集比合理的表达式的完整集合小得多。

A multiplicative correction, being dimensionless, automatically has lower entropy than the additive correction: The set of plausible dimensionless expressions is much smaller than the full set of plausible expressions.

乘法校正为 (x + Δx)(y + Δy)/xy。顾名思义,该比率包含无偿熵。它构造两个有量纲的和 x + Δx 和 y + Δy,将它们相乘,最后将乘积除以 xy。虽然结果是无量纲的,但只有在最后一步才变为无量纲。一种更简洁的方法是通过直接创建无量纲量来对相关因子进行分组:

The multiplicative correction is (x + Δx)(y + Δy)/xy. As written, this ratio contains gratuitous entropy. It constructs two dimensioned sums x + Δx and y + Δy, multiplies them, and finally divides the product by xy. Although the result is dimensionless, it becomes so only in the last step. A cleaner method is to group related factors by making dimensionless quantities right away:

图像

右侧仅由基本无量纲量 1 和有意义的无量纲比率构成:(Δx)/x 是 x 的分数变化,(Δy)/y 是 y 的分数变化。

The right side is built only from the fundamental dimensionless quantity 1 and from meaningful dimensionless ratios: (Δx)/x is the fractional change in x, and (Δy)/y is the fractional change in y.

无谓的熵来自于将 x + Δx、y + Δy、x 和 y 随意混合,它可以通过重组或分离来消除。在物理系统中,分离混合是很困难的。例如,试着将一滴食用色素混入一杯水中。问题是一杯水大约含有 10^ 25个分子。幸运的是,大多数数学表达式的成分较少。我们通常可以重新组合并分离混合的成分,从而降低表达式的熵。

The gratuitous entropy came from mixing x + Δx, y + Δy, x, and y willy nilly, and it was removed by regrouping or unmixing. Unmixing is difficult with physical systems. Try, for example, to remove a drop of food coloring mixed into a glass of water. The problem is that a glass of water contains roughly 1025 molecules. Fortunately, most mathematical expressions have fewer constituents. We can often regroup and unmix the mingled pieces and thereby reduce the entropy of the expression.

问题 5.9 校正因子的矩形

Problem 5.9 Rectangle for the correction factor

画一个矩形代表低熵校正因子

Draw a rectangle representing the low-entropy correction factor

图像

低熵校正因子产生低熵分数变化:

A low-entropy correction factor produces a low-entropy fractional change:

图像

其中 Δ(xy)/xy 是从 xy 到 (x + Δx)(y + Δy) 的分数变化。最右边的项是两个小分数的乘积,因此与前两个项相比较小。如果没有这个小的二次项,

where Δ(xy)/xy is the fractional change from xy to (x + Δx)(y + Δy). The rightmost term is the product of two small fractions, so it is small compared to the preceding two terms. Without this small, quadratic term,

图像

只需添加微小的分数变化即可!

Small fractional changes simply add!

这个分数变化规则比相应的近似规则(绝对变化为 xΔy + yΔx)简单得多。简单性表明熵值较低;事实上,该规则唯一可行的替代方案是分数变化成倍增加的可能性。而这个猜想不太可能成立:当 Δy = 0 时,它预测无论 Δx 的值如何,Δ(xy) = 0(该预测也在问题 5.12中进行了探讨)。

This fractional-change rule is far simpler than the corresponding approximate rule that the absolute change is xΔy + yΔx. Simplicity indicates low entropy; indeed, the only plausible alternative to the proposed rule is the possibility that fractional changes multiply. And this conjecture is not likely: When Δy = 0, it predicts that Δ(xy) = 0 no matter the value of Δx (this prediction is explored also in Problem 5.12).

问题 5.10 热膨胀

Problem 5.10 Thermal expansion

如果由于热膨胀,金属板在各个维度上膨胀 4%,那么它的面积会发生什么变化?

If, due to thermal expansion, a metal sheet expands in each dimension by 4%, what happens to its area?

问题 5.11 折扣涨价

Problem 5.11 Price rise with a discount

想象一下,通货膨胀或版权法导致一本书的价格比去年上涨了10%。幸运的是,作为一名经常买书的人,你开始享受15%的商店折扣。你看到的净价格变化是多少?

Imagine that inflation, or copyright law, increases the price of a book by 10% compared to last year. Fortunately, as a frequent book buyer, you start getting a store discount of 15%. What is the net price change that you see?

5.2.3 平方

5.2.3 Squaring

在分析工程世界和自然世界时,一个常见的运算是平方——乘法的一种特殊情况。长度的平方等于面积,速度的平方与大多数物体所受的阻力成正比(见第 2.4 节):

In analyzing the engineered and natural worlds, a common operation is squaring—a special case of multiplication. Squared lengths are areas, and squared speeds are proportional to the drag on most objects (Section 2.4):

图像

其中,ν 为物体速度,A 为横截面积,ρ 为流体密度。因此,以高速公路速度行驶 d 的距离会消耗能量 E = F d d ∼ ρAv 2 d。因此,降低行驶速度可以降低能耗。在 20 世纪 70 年代油价快速上涨的背景下,这种可能性对西方国家变得尤为重要(参见 [ 7 ] 的分析)。因此,美国将高速公路限速设为 55 英里/小时(90 公里/小时)。

where ν is the speed of the object, A is its cross-sectional area, and ρ is the density of the fluid. As a consequence, driving at highway speeds for a distance d consumes an energy E = Fdd ∼ ρAv2d. Energy consumption can therefore be reduced by driving more slowly. This possibility became important to Western countries in the 1970s when oil prices rose rapidly (see [7] for an analysis). As a result, the United States instituted a highway speed limit of 55 mph (90 kph).

图像 如果以55英里/小时而不是65英里/小时的速度行驶,汽油消耗量会下降多少

By what fraction does gasoline consumption fall due to driving 55 mph instead of 65 mph?

降低限速可以通过降低阻力ρAν 2和缩短行驶距离d来降低油耗:人们更多地根据时间而非距离来衡量和调节通勤。但寻找新住所或新工作是一个缓​​慢的过程。因此,先分析重要事项——在初步分析中假设行驶距离d保持不变(然后尝试解答问题5.14)。

A lower speed limit reduces gasoline consumption by reducing the drag force ρAν2 and by reducing the driving distance d: People measure and regulate their commuting more by time than by distance. But finding a new home or job is a slow process. Therefore, analyze first things first—assume for this initial analysis that the driving distance d stays fixed (then try Problem 5.14).

根据该假设,E 与 ν 2成正比,并且

With that assumption, E is proportional to ν2, and

图像

从65英里/小时加速到55英里/小时,ν大约下降了15%,因此能耗下降了大约30%。高速公路行驶消耗了机动车消耗的石油总量的很大一部分,而在美国,机动车消耗的石油占总石油消耗的很大一部分。因此,30%的降幅大幅降低了美国的石油总消耗量。

Going from 65 mph to 55 mph is roughly a 15% drop in ν, so the energy consumption drops by roughly 30%. Highway driving uses a significant fraction of the oil consumed by motor vehicles, which in the United States consume a significant fraction of all oil consumed. Thus the 30% drop substantially reduced total US oil consumption.

问题 5.12 一个诱人的错误

Problem 5.12 A tempting error

如果 A 和 x 的关系为 A = x 2,那么一个诱人的猜想是

If A and x are related by A = x2, a tempting conjecture is that

图像

使用简单的案例来反驳这个猜想(第 2 章)。

Disprove this conjecture using easy cases (Chapter 2).

问题 5.13 数值估计

Problem 5.13 Numerical estimates

使用分数变化来估计 6.3 3。估计结果有多准确?

Use fractional changes to estimate 6.33. How accurate is the estimate?

问题 5.14 通勤时间限制

Problem 5.14 Time limit on commuting

假设高速公路行驶速度下降15%时,行驶时间(而非行驶距离)保持不变。高速公路行驶所消耗的汽油量会因此发生多少变化?

Assume that driving time, rather than distance, stays fixed as highway driving speeds fall by 15%. What is the resulting fractional change in the gasoline consumed by highway driving?

问题 5.15 风力发电

Problem 5.15 Wind power

理想风力涡轮机的发电量与ν3成正比为什么?)。如果风速仅仅增加10%,对发电量会有什么影响?追求快速风力是人们将风力涡轮机放置在悬崖、山顶或海上的原因之一。

The power generated by an ideal wind turbine is proportional to ν3 (why?). If wind speeds increase by a mere 10%, what is the effect on the generated power? The quest for fast winds is one reason that wind turbines are placed on cliffs or hilltops or at sea.

5.3 具有一般指数的分数变化

5.3 Fractional changes with general exponents

x 2变化的分数变化近似值(第 5.2.3 节)和 x 3变化的分数变化近似值(问题 5.13 )是 x n近似值的特例

The fractional-change approximations for changes in x2 (Section 5.2.3) and in x3 (Problem 5.13) are special cases of the approximation for xn

图像

这条规则提供了一种心算除法(第 5.3.1 节)、估算平方根(第 5.3.2 节)以及判断季节变化的共同解释(第 5.3.3 节)的方法。该规则仅要求分数变化较小且指数 n 不能太大(第 5.3.4 节)。

This rule offers a method for mental division (Section 5.3.1), for estimating square roots (Section 5.3.2), and for judging a common explanation for the seasons (Section 5.3.3). The rule requires only that the fractional change be small and that the exponent n not be too large (Section 5.3.4).

5.3.1 快速心理除法

5.3.1 Rapid mental division

n = −1 的特殊情况提供了快速心算除法的方法。例如,我们估算 1/13。将其重写为 (x + Δx) −1,其中 x = 10,Δx = 3。大的部分是 x −1 = 0.1。由于 (Δx)/x = 30%,因此对 x −1 的小数修正大约为 −30%。结果为 0.07。

The special case n = −1 provides the method for rapid mental division. As an example, let’s estimate 1/13. Rewrite it as (x + Δx)−1 with x = 10 and Δx = 3. The big part is x−1 = 0.1. Because (Δx)/x = 30%, the fractional correction to x−1 is roughly −30%. The result is 0.07.

图像

其中“−30%”符号表示“将前一个对象减少 30%”,是因子 1 − 0.3 的有用简写。

where the “−30%” notation, meaning “decrease the previous object by 30%,” is a useful shorthand for a factor of 1 − 0.3.

图像 估计的准确度如何?误差的来源是什么?

How accurate is the estimate, and what is the source of the error?

估计误差仅为 9%。误差的产生是因为线性近似

The estimate is in error by only 9%. The error arises because the linear approximation

图像

不包括分数变化 (Δx)/x 的平方(或更高次幂)(问题 5.17要求您找到平方项)。

does not include the square (or higher powers) of the fractional change (Δx)/x (Problem 5.17 asks you to find the squared term).

图像 如何减少线性近似中的误差?

How can the error in the linear approximation be reduced?

为了减少误差,可以减小小数部分的变化。由于小数部分的变化取决于大数部分,因此我们来提高大数部分的精度。因此,将 1/13 乘以 8/8(1 的便捷形式),得到 8/104。其大数部分 0.08 已经可以近似于 1/13,误差在 4% 以内。为了改进精度,将 1/104 写为 (x + Δx) −1,其中 x = 100,Δx = 4。小数部分的变化 (Δx)/x 现在为 0.04(而不是 0.3);对 1/x 和 8/x 的小数部分修正仅为 −4%。修正后的估计值为 0.0768:

To reduce the error, reduce the fractional change. Because the fractional change is determined by the big part, let’s increase the accuracy of the big part. Accordingly, multiply 1/13 by 8/8, a convenient form of 1, to construct 8/104. Its big part 0.08 approximates 1/13 already to within 4%. To improve it, write 1/104 as (x + Δx)−1 with x = 100 and Δx = 4. The fractional change (Δx)/x is now 0.04 (rather than 0.3); and the fractional correction to 1/x and 8/x is a mere −4%. The corrected estimate is 0.0768:

图像

这个估算可以在几秒钟内心算完成,并且精确到 0.13%!

This estimate can be done mentally in seconds and is accurate to 0.13%!

问题 5.16 下一个近似值

Problem 5.16 Next approximation

将 1/13 乘以 1 的便捷形式,使分母接近 1000;然后估算 1/13。得到的近似值有多准确?

Multiply 1/13 by a convenient form of 1 to make a denominator near 1000; then estimate 1/13. How accurate is the resulting approximation?

问题 5.17 二次近似

Problem 5.17 Quadratic approximation

求改进分数变分近似中的二次项系数 A

Find A, the coefficient of the quadratic term in the improved fractional-change approximation

图像

使用所得近似值来改进 1/13 的估计值。

Use the resulting approximation to improve the estimates for 1/13.

问题 5.18 燃油效率

Problem 5.18 Fuel efficiency

燃油效率与能耗成反比。如果55英里/小时的限速可以降低30%的能耗,那么之前每加仑汽油可以行驶30英里(每升汽油可以行驶12.8公里)的汽车现在的燃油效率是多少?

Fuel efficiency is inversely proportional to energy consumption. If a 55 mph speed limit decreases energy consumption by 30%, what is the new fuel efficiency of a car that formerly got 30 miles per US gallon (12.8 kilometers per liter)?

5.3.2 平方根

5.3.2 Square roots

分数指数 n = 1/2 提供了估算平方根的方法。例如,我们将图像其重写为 (x + Δx) 1/2,其中 x = 9,Δx = 1。大数部分 x 1/2为 3。由于 (Δx)/x = 1/9 且 n = 1/2,因此分数修正值为 1/18。修正后的估计值为

The fractional exponent n = 1/2 provides the method for estimating square roots. As an example, let’s estimate Rewrite it as (x + Δx)1/2 with x = 9 and Δx = 1. The big part x1/2 is 3. Because (Δx)/x = 1/9 and n = 1/2, the fractional correction is 1/18. The corrected estimate is

图像

精确值为 3.1622...,因此估计精确到 0.14%。

The exact value is 3.1622 . . ., so the estimate is accurate to 0.14%.

问题 5.19 高估还是低估?

Problem 5.19 Overestimate or underestimate?

线性分数变化近似是否会高估所有平方根(因为它高估了图像?如果是,请解释原因;如果不是,请给出反例。

Does the linear fractional-change approximation overestimate all square roots (as it overestimated ? If yes, explain why; if no, give a counterexample.

问题 5.20 余弦近似

Problem 5.20 Cosine approximation

使用小角度近似 sin θ ≈ θ 来证明 cos θ ≈ 1 − θ 2 /2。

Use the small-angle approximation sin θ ≈ θ to show that cos θ ≈ 1 − θ2/2.

问题 5.21 减少零钱

Problem 5.21 Reducing the fractional change

为了减少估计时的分数变化,图像将其重写为图像,然后估计图像结果估计的准确度如何图像

To reduce the fractional change when estimating rewrite it as and then estimate How accurate is the resulting estimate for

问题 5.22 减少分数变化的另一种方法

Problem 5.22 Another method to reduce the fractional change

由于图像与最近的整数平方根相差很小图像,并且图像分数变化无法直接准确地估计。在估计ln 2(第4.3节图像)时也出现了类似的问题;将2重写为(4/3)/(2/3)可以提高精度。这种重写是否有助于估计图像

Because is fractionally distant from the nearest integer square roots and fractional changes do not give a direct and accurate estimate of A similar problem occurred in estimating ln 2 (Section 4.3); there, rewriting 2 as (4/3)/(2/3) improved the accuracy. Does that rewriting help estimate

问题 5.23 立方根

Problem 5.23 Cube root

估计 2 1/3到 10% 以内。

Estimate 21/3 to within 10%.

5.3.3 季节变化的原因是什么?

5.3.3 A reason for the seasons?

人们常说,夏季比冬季温暖,是因为夏季地球比冬季更靠近太阳。这种常见的解释有两个原因,是站不住脚的。首先,尽管南半球与太阳的距离几乎没有差别,但南半球的夏季与北半球的冬季并存。其次,正如我们现在将要估计的那样,变化的地日距离产生的温差太小。距离决定太阳辐射强度,而辐射强度决定地表温度的因果关系,最容易用分形变化来分析。

Summers are warmer than winters, it is often alleged, because the earth is closer to the sun in the summer than in the winter. This common explanation is bogus for two reasons. First, summers in the southern hemisphere happen alongside winters in the northern hemisphere, despite almost no difference in the respective distances to the sun. Second, as we will now estimate, the varying earth–sun distance produces too small a temperature difference. The causal chain—that the distance determines the intensity of solar radiation and that the intensity determines the surface temperature—is most easily analyzed using fractional changes.

太阳辐射强度:强度等于太阳能量除以其辐射面积。太阳能量在一年内几乎不会变化(太阳已经存在了数十亿年);然而,在距离太阳 r 处,能量已经扩散到一个表面积约为 r 2的巨大球体上。因此,强度 I 随 I α r −2变化。半径和强度的分数变化关系如下:

Intensity of solar radiation: The intensity is the solar power divided by the area over which it spreads. The solar power hardly changes over a year (the sun has existed for several billion years); however, at a distance r from the sun, the energy has spread over a giant sphere with surface area ∼ r2. The intensity I therefore varies according to I α r−2. The fractional changes in radius and intensity are related by

图像

地表温度:入射的太阳能无法累积,最终以黑体辐射的形式返回太空。根据斯特藩-玻尔兹曼定律 I = σT 4习题 1.12),其输出强度取决于地球表面温度 T,其中 σ 为斯特藩-玻尔兹曼常数。因此,T α I 1/4。使用分数变化,

Surface temperature: The incoming solar energy cannot accumulate and returns to space as blackbody radiation. Its outgoing intensity depends on the earth’s surface temperature T according to the Stefan–Boltzmann law I = σT4 (Problem 1.12), where σ is the Stefan–Boltzmann constant. Therefore T α I1/4. Using fractional changes,

图像

该关系式将强度与温度联系起来。温度与距离的关系式为 (ΔI)/I = −2 × (Δr)/r。将这两个关系式结合起来,距离与温度的关系如下:

This relation connects intensity and temperature. The temperature and distance are connected by (ΔI)/I = −2 × (Δr)/r. When joined, the two relations connect distance and temperature as follows:

图片

计算的下一步是估算输入值 (Δr)/r,即地日距离的分数变化。地球绕太阳公转的轨道为椭圆形;其轨道距离为

The next step in the computation is to estimate the input (Δr)/r—namely, the fractional change in the earth–sun distance. The earth orbits the sun in an ellipse; its orbital distance is

图片

图片

其中ε是轨道偏心率,θ是极角,l是半直角。因此,r从r min = l/(1 + ε)(当θ = 0°时)变化到r max = l/(1 − ε)(当θ = 180°时)。从r min到l的增加导致了大约ε的微小变化。从l到r max的增加又导致了大约ε的微小变化。因此,r的变化量大约为2ε。对于地球轨道而言,ε = 0.016,因此地日距离变化了0.032或3.2%(导致强度变化6.4%)。

where ε is the eccentricity of the orbit, θ is the polar angle, and l is the semilatus rectum. Thus r varies from rmin = l/(1 + ε) (when θ = 0°) to rmax = l/(1 − ε) (when θ = 180°). The increase from rmin to l contributes a fractional change of roughly ε. The increase from l to rmax contributes another fractional change of roughly ε. Thus, r varies by roughly 2ε. For the earth’s orbit, ε = 0.016, so the earth–sun distance varies by 0.032 or 3.2% (making the intensity vary by 6.4%).

问题 5.24 太阳在哪里?

Problem 5.24 Where is the sun?

上图地球轨道图显示太阳偏离了椭圆的中心。右图则显示了太阳位于另一个或许更自然的位置:椭圆的中心。如果有的话,有哪些物理定律阻止了太阳位于椭圆的中心呢?

The preceding diagram of the earth’s orbit placed the sun away from the center of the ellipse. The diagram to the right shows the sun at an alternative and perhaps more natural location: at the center of the ellipse. What physical laws, if any, prevent the sun from sitting at the center of the ellipse?

图片

问题 5.25 检查分数变化

Problem 5.25 Check the fractional change

查找地球与太阳之间的最小和最大距离,并检查距离是否从最小值到最大值变化了 3.2%。

Look up the minimum and maximum earth–sun distances and check that the distance does vary by 3.2% from minimum to maximum.

距离增加 3.2% 会导致温度略微下降:

A 3.2% increase in distance causes a slight drop in temperature:

图片

然而,人类并不仅仅依靠分数变化而生存,还经历绝对温度变化T。

However, man does not live by fractional changes alone and experiences the absolute temperature change T.

图片

图像 冬季T 0° C ,那么ΔT 0° C

In winter T 0° C, so is ΔT 0° C?

如果我们的计算预测ΔT≈0 ° C,那它肯定是错的。如果用华氏度来测量T,就会得出更不合理的结论,因为在北半球的部分地区,T通常为负值。然而,ΔT不能仅仅因为T是以华氏度来测量的就改变符号!

If our calculation predicts that ΔT 0° C, it must be wrong. An even less plausible conclusion results from measuring T in Fahrenheit degrees, which makes T often negative in parts of the northern hemisphere. Yet ΔT cannot flip its sign just because T is measured in Fahrenheit degrees!

幸运的是,温标受斯特藩-玻尔兹曼定律约束。为了使黑体通量与T 4成正比,温度必须相对于热能为零的状态(即绝对零度)进行测量。摄氏温标和华氏温标均不满足这一要求。

Fortunately, the temperature scale is constrained by the Stefan–Boltzmann law. For blackbody flux to be proportional to T4, temperature must be measured relative to a state with zero thermal energy: absolute zero. Neither the Celsius nor the Fahrenheit scale satisfies this requirement.

相比之下,开尔文温标测量的是相对于绝对零度的温度。在开尔文温标中,平均地表温度 T 300 K;因此,T 变化 1.6% 会导致 ΔT 5 K。5 K 的变化也意味着 5°C 的变化——开尔文和摄氏度大小相同,尽管这两个温标的零点不同。(另见问题 5.26。)温带地区夏季和冬季之间的典型温差为 20°C——即使考虑到估算中的误差,也远大于预测的 5°C 变化。日地距离变化对于季节变化的原因是一个令人怀疑的解释。

In contrast, the Kelvin scale does measure temperature relative to absolute zero. On the Kelvin scale, the average surface temperature is T 300 K; thus, a 1.6% change in T makes ΔT 5 K. A 5 K change is also a 5° C change—Kelvin and Celsius degrees are the same size, although the scales have different zero points. (See also Problem 5.26.) A typical temperature change between summer and winter in temperate latitudes is 20° C—much larger than the predicted 5° C change, even after allowing for errors in the estimate. A varying earth–sun distance is a dubious explanation of the reason for the seasons.

问题 5.26 转换为华氏度

Problem 5.26 Converting to Fahrenheit

华氏温度和摄氏温度之间的转换是

The conversion between Fahrenheit and Celsius temperatures is

图片

所以,5°C 的变化应该相当于 41°F 的变化——这足以解释季节变化了!这个推理有什么问题?

so a change of 5° C should be a change of 41° F—sufficiently large to explain the seasons! What is wrong with this reasoning?

问题 5.27 替代解释

Problem 5.27 Alternative explanation

如果距离太阳的距离变化都无法解释季节变化,那什么可以呢?你的提案应该顺便解释一下,为什么南北半球的夏季相差6个月。

If a varying distance to the sun cannot explain the seasons, what can? Your proposal should, in passing, explain why the northern and southern hemispheres have summer 6 months apart.

5.3.4 有效期限制

5.3.4 Limits of validity

线性分数变化近似

The linear fractional-change approximation

图片

一直很有用。但它何时有效?为了避免被符号淹没,我们可以将 Δx 写为 z;然后取 x = 1,使 z 为绝对值和分数变化。右边变成 nz,线性分数变化近似等价于

has been useful. But when is it valid? To investigate without drowning in notation, write z for Δx; then choose x = 1 to make z the absolute and the fractional change. The right side becomes nz, and the linear fractional-change approximation is equivalent to

图片

当 z 过大时,近似值会变得不准确:例如,当图片z = 1 时(问题 5.22)。指数 n 是否也受到限制?前面的例子仅展示了中等大小的指数:n = 2 表示能耗(第 5.2.3 节),-2 表示燃油效率(问题 5.18),-1 表示倒数(第 5.3.1 节),1/2 表示平方根(第 5.3.2 节),-2 和 1/4 表示季节(第 5.3.3 节)。我们需要更多数据。

The approximation becomes inaccurate when z is too large: for example, when evaluating with z = 1 (Problem 5.22). Is the exponent n also restricted? The preceding examples illustrated only moderate-sized exponents: n = 2 for energy consumption (Section 5.2.3), −2 for fuel efficiency (Problem 5.18), −1 for reciprocals (Section 5.3.1), 1/2 for square roots (Section 5.3.2), and −2 and 1/4 for the seasons (Section 5.3.3). We need further data.

图像 在指数较大的极端情况下会发生什么?

What happens in the extreme case of large exponents?

对于较大的指数,例如 n = 100 和 z = 0.001,近似值预测 1.001 100 ≈ 1.1——接近真实值 1.105……然而,选择相同的 n 和 z = 0.1(大于 0.001 但仍然很小)会产生糟糕的预测

With a large exponent such as n = 100 and, say, z = 0.001, the approximation predicts that 1.001100 ≈ 1.1—close to the true value of 1.105 . . . However, choosing the same n alongside z = 0.1 (larger than 0.001 but still small) produces the terrible prediction

图片

1.1 100大约是 14,000,比预测值大 1000 多倍。

1.1100 is roughly 14,000, more than 1000 times larger than the prediction.

两个预测都使用了较大的 n 和较小的 z,但只有一个预测准确;因此,问题不可能仅仅出在 n 或 z 上。罪魁祸首或许是无量纲乘积 nz。为了验证这一想法,请保持 nz 不变,同时尝试较大的 n 值。对于 nz 来说,合理的常数是 1——最简单的无量纲数。以下是几个例子。

Both predictions used large n and small z, yet only one prediction was accurate; thus, the problem cannot lie in n or z alone. Perhaps the culprit is the dimensionless product nz. To test that idea, hold nz constant while trying large values of n. For nz, a sensible constant is 1—the simplest dimensionless number. Here are several examples.

图片

在每个例子中,近似值都错误地预测 (1 + z) n = 2。

In each example, the approximation incorrectly predicts that (1 + z)n = 2.

图像 错误原因是什么?

What is the cause of the error?

为了找到原因,将序列延伸至 1.001 1000以上,并希望出现一种模式:这些值似乎接近于 e = 2.718281828……(自然对数的底数)。因此,对整个近似值取对数。

To find the cause, continue the sequence beyond 1.0011000 and hope that a pattern will emerge: The values seem to approach e = 2.718281828 . . ., the base of the natural logarithms. Therefore, take the logarithm of the whole approximation.

图片

图片

图形推理表明,当 z 1 时,ln(1 + z) z 图像第 4.3 节)。因此,n ln(1 + z) nz,使得 (1 + z) n ≈ e nz。这种改进的近似值解释了为什么近似值 (1 + z) n ≈ 1 + nz 在 nz 较大时失效:只有当 nz 图像1 大约等于 e nz 1 + nz 时才成立。因此,当 z 图像1 时,两个最简单的近似值是

Pictorial reasoning showed that ln(1 + z) z when z 1 (Section 4.3). Thus, n ln(1 + z) nz, making (1 + z)n ≈ enz. This improved approximation explains why the approximation (1 + z)n ≈ 1 + nz failed with large nz: Only when nz 1 is enz approximately 1 + nz. Therefore, when z 1 the two simplest approximation are

图片

该图显示了整个 n-z 平面上每个区域的最简近似。坐标轴为对数坐标,n 和 z 假设为正:右半平面 z 为图像1,上半平面 n 为图像1。右下角的边界曲线为 n ln z = 1。解释边界并扩展近似值是一项有益的练习(习题 5.28)。

The diagram shows, across the whole n–z plane, the simplest approximation in each region. The axes are logarithmic and n and z are assumed positive: The right half plane shows z 1, and the upper half plane shows n 1. On the lower right, the boundary curve is n ln z = 1. Explaining the boundaries and extending the approximations is an instructive exercise (Problem 5.28).

图片

问题 5.28 解释近似平面

Problem 5.28 Explaining the approximation plane

在右半平面中,解释 n/z = 1 和 n ln z = 1 边界。对于整个平面,尽可能放宽 n 和 z 为正的假设。

In the right half plane, explain the n/z = 1 and n ln z = 1 boundaries. For the whole plane, relax the assumption of positive n and z as far as possible.

问题 5.29 二项式定理推导

Problem 5.29 Binomial-theorem derivation

尝试以下 (1+z) n ≈ e nz(其中 n ≥ 图像1)的另一种推导方法。利用二项式定理展开 (1 + z) n ,将 n − k 近似为 n ,简化二项式系数的乘积,并将所得展开式与 e nz的泰勒级数进行比较。

Try the following alternative derivation of (1+z)n ≈ enz (where n 1). Expand (1 + z)n using the binomial theorem, simplify the products in the binomial coefficients by approximating n − k as n, and compare the resulting expansion to the Taylor series for enz.

5.4 逐次逼近:井有多深?

5.4 Successive approximation: How deep is the well?

下一个取出大部分的说明强调了逐次逼近,并伪装成物理问题。

The next illustration of taking out the big part emphasizes successive approximation and is disguised as a physics problem.

你把一块石头扔进一口深度未知的井里,4秒后听到了水花飞溅的声音。忽略空气阻力,求出h,误差在5%以内。设c s = 340 ms −1为声速,g = 10 ms −2为重力强度。

You drop a stone down a well of unknown depth h and hear the splash 4 s later. Neglecting air resistance, find h to within 5%. Use cs = 340 ms−1 as the speed of sound and g = 10 ms−2 as the strength of gravity.

近似解和精确解给出了几乎相同的井深,但却提供了截然不同的理解。

Approximate and exact solutions give almost the same well depth, but offer significantly different understandings.

5.4.1 精确深度

5.4.1 Exact depth

深度由以下约束决定:4 秒的等待时间分为两个时间:岩石自由落下井的时间和声音沿井向上传播的时间。自由落体时间为图像问题 1.3),因此总时间为

The depth is determined by the constraint that the 4 s wait splits into two times: the rock falling freely down the well and the sound traveling up the well. The free-fall time is (Problem 1.3), so the total time is

图片

为了精确求解 h,要么将平方根放在一边,然后对两边求平方,得到 h 的二次方程(问题 5.30);或者,为了减少错误,将约束重写为新变量 z = 的二次方程图片

To solve for h exactly, either isolate the square root on one side and square both sides to get a quadratic equation in h (Problem 5.30); or, for a less error-prone method, rewrite the constraint as a quadratic equation in a new variable z = .

问题 5.30 其他二次方程

Problem 5.30 Other quadratic

通过分离一侧的平方根并同时求两侧的平方来求解h。与将约束重写为z = 的二次函数相比,此方法的优缺点是什么图片

Solve for h by isolating the square root on one side and squaring both sides. What are the advantages and disadvantages of this method in comparison with the method of rewriting the constraint as a quadratic in z = ?

作为 z = 的二次方程图片,约束为

As a quadratic equation in z = , the constraint is

图片

使用二次公式并选择正根可得出

Using the quadratic formula and choosing the positive root yields

图片

因为 z 2 = h,

Because z2 = h,

图像

代入 g = 10 ms −2和 c s = 340 ms −1得出 h ≈ 71.56 m。

Substituting g = 10 ms−2 and cs = 340 ms−1 gives h ≈ 71.56 m.

即使深度正确,其精确公式也一团糟。这种高熵的恐怖经常源于二次公式;它的使用往往标志着符号操控战胜了思维。我们会发现,精确答案可能不如近似答案有用。

Even if the depth is correct, the exact formula for it is a mess. Such high-entropy horrors arise frequently from the quadratic formula; its use often signals the triumph of symbol manipulation over thought. Exact answers, we will find, may be less useful than approximate answers.

5.4.2 大致深度

5.4.2 Approximate depth

要找到低熵的近似深度,需要确定最重要的部分——最重要的效应。这里,大部分时间是岩石的自由落体:即使岩石自由落体持续了4秒,其最大速度也只有gΤ = 40 ms −1,远低于c s。因此,最重要的效应应该出现在声速无限大的极端情况下。

To find a low-entropy, approximate depth, identify the big part—the most important effect. Here, most of the total time is the rock’s free fall: The rock’s maximum speed, even if it fell for the entire 4 s, is only gΤ = 40 ms−1, which is far below cs. Therefore, the most important effect should arise in the extreme case of infinite sound speed.

图像 如果c s = ∞,那么井有多深?

If cs = ∞, how deep is the well?

在这个零阶近似中,自由落体时间 t 0为全时间 T = 4 s,因此井深 h 0变为

In this zeroth approximation, the free-fall time t0 is the full time T = 4 s, so the well depth h0 becomes

图像

图像 这个近似深度是高估了还是低估了?它的准确度如何?

Is this approximate depth an overestimate or underestimate? How accurate is it?

这种近似值忽略了声音的传播时间,因此高估了自由落体的时间,从而也高估了深度。与大约71.56米的真实深度相比,它仅高估了11%——对于一种能够提供物理洞察力的快速方法来说,这个精度还算合理。此外,这种近似值本身也体现了其改进之处。

This approximation neglects the sound-travel time, so it overestimates the free-fall time and therefore the depth. Compared to the true depth of roughly 71.56 m, it overestimates the depth by only 11%—reasonable accuracy for a quick method offering physical insight. Furthermore, this approximation suggests its own refinement.

图像 如何改进这种近似?

How can this approximation be improved?

为了改进它,使用近似深度 h 0来近似声音传播时间。

To improve it, use the approximate depth h0 to approximate the sound-travel time.

图像

图像

剩余时间是自由落体时间的下一个近似值。

The remaining time is the next approximation to the free-fall time.

图像

在这段时间内,岩石下落了一段距离图像,因此深度的下一个近似值是

In that time, the rock falls a distance so the next approximation to the depth is

图像

图像 这个近似深度是高估了还是低估了?它的准确度如何?

Is this approximate depth an overestimate or underestimate? How accurate is it?

h 1的计算使用了 h 0来估算声音的传播时间。由于 h 0高估了深度,因此该程序会高估声音的传播时间,同时低估自由落体时间,两者之差相同。因此,h 1低估了深度。事实上,h 1略小于大约 71.56 米的真实深度,但仅小了 1.3%。

The calculation of h1 used h0 to estimate the sound-travel time. Because h0 overestimates the depth, the procedure overestimates the sound-travel time and, by the same amount, underestimates the free-fall time. Thus h1 underestimates the depth. Indeed, h1 is slightly smaller than the true depth of roughly 71.56 m—but by only 1.3%.

与精确求解二次公式相比,逐次逼近法有几个优点。首先,它有助于我们加深对系统的物理理解;例如,我们意识到T = 4秒的大部分时间都花在了自由落体上,因此深度大约为gT² / 2。其次,它有图形化的解释(问题5.34)。第三,它能快速给出足够准确的答案。如果你想知道跳进井里是否安全,为什么要把深度计算到小数点后三位?

The method of successive approximation has several advantages over solving the quadratic formula exactly. First, it helps us develop a physical understanding of the system; we realize, for example, that most of the T = 4 s is spent in free fall, so the depth is roughly gT2/2. Second, it has a pictorial explanation (Problem 5.34). Third, it gives a suffciently accurate answer quickly. If you want to know whether it is safe to jump into the well, why calculate the depth to three decimal places?

最后,该方法可以处理模型中的微小变化。例如,声速可能随深度变化,或者空气阻力变得很重要(问题 5.32)。这时,强力的二次公式方法就失效了。二次公式,甚至更复杂的三次和四次公式,都是复杂方程的罕见闭式解。大多数方程没有闭式解。因此,如果我们要求一个精确的答案,对可解模型进行微小的改动通常就会产生一个难以处理的模型。逐次逼近法是一种稳健的替代方法,可以产生低熵、易于理解的解。

Finally, the method can handle small changes in the model. Maybe the speed of sound varies with depth, or air resistance becomes important (Problem 5.32). Then the brute-force, quadratic-formula method fails. The quadratic formula and the even messier cubic and the quartic formulas are rare closed-form solutions to complicated equations. Most equations have no closed-form solution. Therefore, a small change to a solvable model usually produces an intractable model—if we demand an exact answer. The method of successive approximation is a robust alternative that produces low-entropy, comprehensible solutions.

问题 5.31 参数值不准确

Problem 5.31 Parameter-value inaccuracies

深度的二阶近似值 h 2是多少?将 h 1和 h 2的误差与使用 g = 10 ms −2产生的误差进行比较。

What is h2, the second approximation to the depth? Compare the error in h1 and h2 with the error made by using g = 10 ms−2.

问题 5.32 空气阻力的影响

Problem 5.32 Effect of air resistance

忽略空气阻力(第 2.4.2 节)会导致深度产生大约多少分数的误差?将此误差与一阶近似值 h 1和二阶近似值 h 2 的误差(问题 5.31)进行比较。

Roughly what fractional error in the depth is produced by neglecting air resistance (Section 2.4.2)? Compare this error to the error in the first approximation h1 and in the second approximation h2 (Problem 5.31).

问题 5.33 井深分析的无量纲形式

Problem 5.33 Dimensionless form of the well-depth analysis

即使是最混乱的结果,在无量纲形式下也更清晰,熵也更低。h、g、T 和 c 四个量产生两个独立的无量纲组(第 2.4.1 节)。直观上合理的一对是

Even the messiest results are cleaner and have lower entropy in dimensionless form. The four quantities h, g, T, and cs produce two independent dimensionless groups (Section 2.4.1). An intuitively reasonable pair are

图像

a. 的物理解释是什么图像

a. What is a physical interpretation of ?

b. 对于两个群,一般无量纲形式为图像= f( )。简单情况下图像结果为 0 吗?图像图像

b. With two groups, the general dimensionless form is = f(). What is in the easy case 0?

c. 重写二次公式解

c. Rewrite the quadratic-formula solution

图像

as 图像= f( 图像)。然后检查 f( 图像) 在简单情况下是否正确运行图像 0。

as = f(). Then check that f() behaves correctly in the easy case 0.

问题 5.34 井深的时空图

Problem 5.34 Spacetime diagram of the well depth

时空图 [ 44 ]如何表示井深的逐次近似?在图上,标记 h 0(深度的零次近似值)、h 1和精确深度 h。标记 t 0,即自由落体时间的零次近似值。为什么岩石和声波前曲线的部分是用虚线表示的?如果声速加倍,你会如何重新绘制该图?如果 g 加倍?

How does the spacetime diagram [44] illustrate the successive approximation of the well depth? On the diagram, mark h0 (the zeroth approximation to the depth), h1, and the exact depth h. Mark t0, the zeroth approximation to the free-fall time. Why are portions of the rock and sound-wavefront curves dotted? How would you redraw the diagram if the speed of sound doubled? If g doubled?

图像

5.5 令人畏惧的三角积分

5.5 Daunting trigonometric integral

最后一个例子是估算一个令人望而生畏的三角积分,这是我本科时学过的。我和同学们经常在物理图书馆熬夜解答作业;研究生们也一样,用他们最喜欢的数学和物理问题来招待我们。

The final example of taking out the big part is to estimate a daunting trigonometric integral that I learned as an undergraduate. My classmates and I spent many late nights in the physics library solving homework problems; the graduate students, doing the same for their courses, would regale us with their favorite mathematics and physics problems.

这个积分出现在前苏联朗道理论物理研究所的数学预科考试中。题目要求计算

The integral appeared on the mathematical-preliminaries exam to enter the Landau Institute for Theoretical Physics in the former USSR. The problem is to evaluate

图像

无需使用计算器或计算机,在不到 5 分钟的时间内即可达到 5% 以内!

to within 5% in less than 5 min without using a calculator or computer!

(cos t) 100看起来吓人。大多数三角恒等式都帮不上忙。通常有用的恒等式 (cos t) 2 = (cos 2t − 1)/2 只会产生

That (cos t)100 looks frightening. Most trigonometric identities do not help. The usually helpful identity (cos t)2 = (cos 2t − 1)/2 produces only

图像

它在展开 50 次方后就变成了一个三角函数怪物。

which becomes a trigonometric monster upon expanding the 50th power.

一个指向更简单方法的线索是,5% 的精度就足够了——所以,找到最大的部分!当 t 接近于零时,被积函数最大。此时,成本 t 1 − t 2 /2 (问题 5.20),因此被积函数大致为

A clue pointing to a simpler method is that 5% accuracy is sufficient—so, find the big part! The integrand is largest when t is near zero. There, cos t 1 − t2/2 (Problem 5.20), so the integrand is roughly

图像

它具有我们熟悉的形式 (1 + z) n,其中小数变化 z = −t 2 /2 ,指数 n = 100。当 t 很小时,z = −t 2 /2 很小,因此 (1 + z) n可以使用第 5.3.4 节的结果来近似:

It has the familiar form (1 + z)n, with fractional change z = −t2/2 and exponent n = 100. When t is small, z = −t2/2 is tiny, so (1 + z)n may be approximated using the results of Section 5.3.4:

图像

由于指数 n 很大,即使 t 和 z 较小,nz 也可能很大。因此,最安全的近似值为 (1 + z) n ≈ e nz;则

Because the exponent n is large, nz can be large even when t and z are small. Therefore, the safest approximation is (1 + z)n ≈ enz; then

图像

余弦的高次幂变为高斯!

A cosine raised to a high power becomes a Gaussian!

为了检验这一令人惊讶的结论,计算机生成的 (cost t) n (n = 1 ... 5) 图显示,随着 n 的增加,形成了高斯钟形。

As a check on this surprising conclusion, computer-generated plots of (cos t)n for n = 1 . . . 5 show a Gaussian bell shape taking form as n increases.

即使有这样的图形证据,用高斯函数代替 (cost t) 100也有点可疑。在原始积分中,t 的范围从 −π/2 到 π/2,而这些端点远远超出了 cost t 1 − t 2 /2 的精确近似值区域。幸运的是,这个问题只会导致一个很小的误差(问题 5.35)。忽略这个误差会将原始积分变成具有有限极限的高斯积分:

Even with this graphical evidence, replacing (cos t)100 by a Gaussian is a bit suspicious. In the original integral, t ranges from −π/2 to π/2, and these endpoints are far outside the region where cos t 1 − t2/2 is an accurate approximation. Fortunately, this issue contributes only a tiny error (Problem 5.35). Ignoring this error turns the original integral into a Gaussian integral with finite limits:

图像

图像

不幸的是,在有限极限下,积分没有闭合形式。但将极限扩展到无穷大可以得到一个闭合形式,而且几乎不产生误差(问题 5.36)。近似链现在是

Unfortunately, with finite limits the integral has no closed form. But extending the limits to infinity produces a closed form while contributing almost no error (Problem 5.36). The approximation chain is now

图像

问题 5.35 使用原始限制

Problem 5.35 Using the original limits

近似值成本 t 1 − t 2 /2 要求 t 较小。为什么在小 t 范围之外使用近似值不会产生显著误差?

The approximation cos t 1 − t2/2 requires that t be small. Why doesn’t using the approximation outside the small-t range contribute a significant error?

问题 5.36 扩展极限

Problem 5.36 Extending the limits

为什么将积分极限从± π/2 扩展到±∞不会导致重大误差?

Why doesn’t extending the integration limits from ±π/2 to ±∞ contribute a significant error?

最后一个积分是老朋友了(第 2.1 节):图像= 图像。当 α = 50 时,积分变为图像。方便的是,50 大约等于 16π,因此平方根(也就是我们 5% 的估计值)大约等于 0.25。

The last integral is an old friend (Section 2.1): = . With α = 50, the integral becomes . Conveniently, 50 is roughly 16π, so the square root—and our 5% estimate—is roughly 0.25.

为了比较,精确积分是(问题 5.41

For comparison, the exact integral is (Problem 5.41)

图像

当 n = 100 时,二项式系数和二的幂产生

When n = 100, the binomial coeffcient and power of two produce

图像

我们对 5 分钟内 5% 以内的估计为 0.25,精确到几乎 0.01%!

Our 5-minute, within-5% estimate of 0.25 is accurate to almost 0.01%!

问题 5.37 绘制近似值

Problem 5.37 Sketching the approximations

绘制 (cost t) 100及其两个近似值 e −50t2和 1 − 50t 2

Plot (cos t)100 and its two approximations e−50t2 and 1 − 50t2.

问题 5.38 最简单的近似值

Problem 5.38 Simplest approximation

使用线性分数阶近似 (1 − t 2 /2) 100 ≈ 1 − 50t 2来近似被积函数;然后在 1 − 50t 2为正的范围内对其进行积分。这个 1 分钟方法的结果与精确值 0.2500 有多接近?

Use the linear fractional-change approximation (1 − t2/2)100 ≈ 1 − 50t2 to approximate the integrand; then integrate it over the range where 1 − 50t2 is positive. How close is the result of this 1-minute method to the exact value 0.2500 . . .?

问题 5.39 巨大指数

Problem 5.39 Huge exponent

估计

Estimate

图像

问题 5.40 你能走多低?

Problem 5.40 How low can you go?

调查近似值的准确性

Investigate the accuracy of the approximation

图像

对于较小的 n,包括 n = 1。

for small n, including n = 1.

问题 5.41 闭式

Problem 5.41 Closed form

评估积分

To evaluate the integral

图像

以封闭形式,使用以下步骤:

in closed form, use the following steps:

a. 将 cost t 替换为 (e it + e −it )/2。

a. Replace cos t with (eit + e−it)/2.

b.利用二项式定理展开100次方。

b. Use the binomial theorem to expand the 100th power.

c. 将像 e ikt这样的每一项与其对应的 e −ikt配对;然后对它们的和从 −π/2 到 π/2 进行积分。k 的哪个或哪些值能使和的积分不为零?

c. Pair each term like eikt with a counterpart e−ikt; then integrate their sum from −π/2 to π/2. What value or values of k produce a sum whose integral is nonzero?

5.6 总结和进一步的问题

5.6 Summary and further problems

遇到复杂问题时,将其分解成一个主要部分(最重要的影响)和一个修正部分。先分析主要部分,然后再考虑修正部分。这种逐次逼近法,一种分而治之的推理,会自动以低熵形式给出结果。低熵表达式几乎没有合理的替代方案;因此,它们更容易记忆和理解。简而言之,近似结果可能比精确结果更有用。

Upon meeting a complicated problem, divide it into a big part—the most important effect—and a correction. Analyze the big part first, and worry about the correction afterward. This successive-approximation approach, a species of divide-and-conquer reasoning, gives results automatically in a low-entropy form. Low-entropy expressions admit few plausible alternatives; they are therefore memorable and comprehensible. In short, approximate results can be more useful than exact results.

问题 5.42 大对数

Problem 5.42 Large logarithm

ln(1+e 2 )中最大的部分是什么?给出一个简短的计算,将ln(1+e 2 )的精度控制在2%以内。

What is the big part in ln(1+e2)? Give a short calculation to estimate ln(1+e2) to within 2%.

问题 5.43 细菌突变

Problem 5.43 Bacterial mutations

20世纪90年代,加州理工学院的一次生物学研讨会上描述了一项实验。研究人员反复照射一群细菌,以产生突变。在每一轮照射中,5%的细菌会发生突变。经过140轮照射后,大约有多少比例的细菌没有发生突变?(研讨会演讲者给观众3秒钟的时间进行猜测,这段时间几乎不够使用甚至找到计算器。)

In an experiment described in a Caltech biology seminar in the 1990s, researchers repeatedly irradiated a population of bacteria in order to generate mutations. In each round of radiation, 5% of the bacteria got mutated. After 140 rounds, roughly what fraction of bacteria were left unmutated? (The seminar speaker gave the audience 3 s to make a guess, hardly enough time to use or even find a calculator.)

问题 5.44 二次方程重温

Problem 5.44 Quadratic equations revisited

下面的二次方程受到[ 29 ]的启发,描述了一个阻尼非常强的振荡系统。

The following quadratic equation, inspired by [29], describes a very strongly damped oscillating system.

图像

a. 使用二次公式和标准计算器求二次方程的两个根。哪里出错了?为什么?

a. Use the quadratic formula and a standard calculator to find both roots of the quadratic. What goes wrong and why?

b. 通过去除较大部分来估算根。(提示:在适当的极端情况下,近似并求解方程。)然后使用逐次逼近法改进估计值。

b. Estimate the roots by taking out the big part. (Hint: Approximate and solve the equation in appropriate extreme cases.) Then improve the estimates using successive approximation.

c. 二次公式分析与逐次逼近法相比有哪些优缺点?

c. What are the advantages and disadvantages of the quadratic-formula analysis versus successive approximation?

问题 5.45 二项分布的正态近似

Problem 5.45 Normal approximation to the binomial distribution

二项式展开

The binomial expansion

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包含以下形式的术语

contains terms of the form

图像

其中 k = −n ... n。每一项 f(k) 表示在 2n 次抛硬币中,抛出 n − k 次正面(和 n + k 次反面)的概率;f(k) 是所谓的二项分布,其参数为 p = q = 1/2。通过回答以下问题来近似计算该分布:

where k = −n . . . n. Each term f(k) is the probability of tossing n − k heads (and n + k tails) in 2n coin flips; f(k) is the so-called binomial distribution with parameters p = q = 1/2. Approximate this distribution by answering the following questions:

a. f(k)是k的偶函数还是奇函数?当k为多少时,f(k)有最大值?

a. Is f(k) an even or an odd function of k? For what k does f(k) have its maximum?

b. 当 k « n 时,近似 f(k),并画出 f(k) 的简图。由此推导并解释二项分布的正态近似。

b. Approximate f(k) when k « n and sketch f(k). Therefore, derive and explain the normal approximation to the binomial distribution.

c. 用正态近似法证明该二项分布的方差为n/2。

c. Use the normal approximation to show that the variance of this binomial distribution is n/2.

问题 5.46 Beta 函数

Problem 5.46 Beta function

贝叶斯推理中经常出现以下积分:

The following integral appears often in Bayesian inference:

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其中 f(a − 1, b − 1) 是欧拉贝塔函数。使用街头斗争法推测 f(a, 0)、f(a, a) 以及最终 f(a, b) 的函数形式。使用高质量的积分表或 Maxima 等计算机代数系统来验证你的推测。

where f(a − 1, b − 1) is the Euler beta function. Use street-fighting methods to conjecture functional forms for f(a, 0), f(a, a), and, finally, f(a, b). Check your conjectures with a high-quality table of integrals or a computer-algebra system such as Maxima.

6

6

类比

Analogy

处境艰难时,强者会降低标准。这一理念,也就是整本书的主题,是最后一个街头格斗工具——类比推理的基础。它的建议很简单:面对一个难题,构造并解决一个类似但更简单的问题——一个类比问题。熟能生巧。该工具在空间三角学(第 6.1 节)中引入;在立体几何和拓扑学(第 6.2 节)中得到进一步完善;然后应用于离散数学(第 6.3 节),并在最后的例子中应用于无穷超越和(第 6.4 节)。

When the going gets tough, the tough lower their standards. This idea, the theme of the whole book, underlies the final street-fighting tool of reasoning by analogy. Its advice is simple: Faced with a difficult problem, construct and solve a similar but simpler problem—an analogous problem. Practice develops fluency. The tool is introduced in spatial trigonometry (Section 6.1); sharpened on solid geometry and topology (Section 6.2); then applied to discrete mathematics (Section 6.3) and, in the farewell example, to an infinite transcendental sum (Section 6.4).

6.1 空间三角学:甲烷中的键角

6.1 Spatial trigonometry: The bond angle in methane

第一个类比来自空间三角学。在甲烷(化学式 CH4 中,一个碳原子位于正四面体的中心,每个顶点各有一个氢原子。两个碳氢键之间的夹角 θ 是多少?

The first analogy comes from spatial trigonometry. In methane (chemical formula CH4), a carbon atom sits at the center of a regular tetrahedron, and one hydrogen atom sits at each vertex. What is the angle θ between two carbon–hydrogen bonds?

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三维空间中的角度很难直观地表示。例如,试着想象并计算正四面体两个面之间的角度。由于二维角度很容易直观地表示,让我们构建并分析一个类似的平面分子。了解它的键角或许有助于我们猜测甲烷的键角。

Angles in three dimensions are hard to visualize. Try, for example, to imagine and calculate the angle between two faces of a regular tetrahedron. Because two-dimensional angles are easy to visualize, let’s construct and analyze an analogous planar molecule. Knowing its bond angle might help us guess methane’s bond angle.

图像 类似的平面分子应该有四个还是三个氢原子?

Should the analogous planar molecule have four or three hydrogens?

四个氢原子形成四个键,当它们在平面上规则排列时,会产生两个不同的键角。相比之下,甲烷只有一个键角。因此,使用四个氢原子会改变原始问题的一个关键特征。可能的解决办法是构建类似的仅使用三个氢原子的平面分子。

Four hydrogens produce four bonds which, when spaced regularly in a plane, produce two different bond angles. In contrast, methane contains only one bond angle. Therefore, using four hydrogens alters a crucial feature of the original problem. The likely solution is to construct the analogous planar molecule using only three hydrogens.

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三个氢原子规则地排列在一个平面上,只会产生一个键角:θ = 120°。或许这个键角就是甲烷的键角!然而,一个数据点对于更高维度的预测来说,只是纸上谈兵。二维空间(d = 2)的单个数据点与许多猜想相符——例如,在d维空间中,键角是120°、(60d)°,或者许多其他值。

Three hydrogens arranged regularly in a plane create only one bond angle: θ = 120°. Perhaps this angle is the bond angle in methane! One data point, however, is a thin reed on which to hang a prediction for higher dimensions. The single data point for two dimensions (d = 2) is consistent with numerous conjectures—for example, that in d dimensions the bond angle is 120° or (60d)° or much else.

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选择一个合理的猜想需要收集更多数据。一个更简单但更类似的问题很容易获得数据:一维线性分子 CH2 它的两个氢原子彼此相对,因此两个 C–H 键形成 θ = 180° 的夹角。

Selecting a reasonable conjecture requires gathering further data. Easily available data comes from an even simpler yet analogous problem: the one-dimensional, linear molecule CH2. Its two hydrogens sit opposite one another, so the two C–H bonds form an angle of θ = 180°.

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图像 根据积累的数据,对于三维角度θ 3有哪些合理的推测?

Based on the accumulated data, what are reasonable conjectures for the three-dimensional angle θ3?

一维分子排除了θ d = (60d)°的猜想。它还提出了新的猜想,例如θ d = (240 − 60d)°或θ d = 360°/(d + 1)。检验这些猜想是简单情况方法的理想任务。高维(高维)简单情况检验推翻了θ d = (240 − 60d)°的猜想。对于高维分子,它预测了不合理的键角,即当d = 4时θ = 0,当d > 4时θ < 0。

The one-dimensional molecule eliminates the conjecture that θd = (60d)°. It also suggests new conjectures—for example, that θd = (240 − 60d)° or θd = 360°/(d + 1). Testing these conjectures is an ideal task for the method of easy cases. The easy-cases test of higher dimensions (high d) refutes the conjecture that θd = (240 − 60d)°. For high d, it predicts implausible bond angles—namely, θ = 0 for d = 4 and θ < 0 for d > 4.

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幸运的是,第二个假设 θ d = 360°/(d + 1) 通过了同样的简单情况检验。让我们继续验证它对甲烷的预测——即 θ 3 = 90°。想象一下甲烷的一个“大哥”:一个 CH 6分子,碳原子位于立方体的中心,六个氢原子位于立方体的中心。它的小键角为 90°。(另一个键角为 180°。)现在移除两个氢原子,将 CH 6变成 CH 4 ,使剩下的四个氢原子均匀分布。减少拥挤度会使小键角超过 90°,从而推翻 θ 3 = 90°的预测。

Fortunately, the second suggestion, θd = 360°/(d + 1), passes the same easy-cases test. Let’s continue to test it by evaluating its prediction for methane—namely, θ3 = 90°. Imagine then a big brother of methane: a CH6 molecule with carbon at the center of a cube and six hydrogens at the face centers. Its small bond angle is 90°. (The other bond angle is 180°.) Now remove two hydrogens to turn CH6 into CH4, evenly spreading out the remaining four hydrogens. Reducing the crowding raises the small bond angle above 90°—and refutes the prediction that θ3 = 90°.

问题 6.1 有多少个氢?

Problem 6.1 How many hydrogens?

在类似的四维和五维键角问题中,需要多少个氢原子?利用这些信息证明θ 4 > 90°。对于所有d, θ d > 90°吗?

How many hydrogens are needed in the analogous four-and five-dimensional bond-angle problems? Use this information to show that θ4 > 90°. Is θd > 90° for all d?

迄今为止的数据已经推翻了最简单的有理函数猜想 (240−60d)° 和 360°/(d+1)。虽然其他有理函数猜想可能仍然存在,但只有两个数据点,可能性实在太大。更糟糕的是,θ d甚至可能不是 d 的有理函数。

The data so far have refuted the simplest rational-function conjectures (240−60d)° and 360°/(d+1). Although other rational-function conjectures might survive, with only two data points the possibilities are too vast. Worse, θd might not even be a rational function of d.

进步需要新的思路:键角或许并非最容易研究的变量。在推测3、5、11、29……系列中的下一个项时,也会出现类似的困难。

Progress requires a new idea: The bond angle might not be the simplest variable to study. An analogous difficulty arises when conjecturing the next term in the series 3, 5, 11, 29, . . .

图像 该系列的下一个术语是什么?

What is the next term in the series?

乍一看,这些数字似乎几乎是随机的。然而,从每一项中减去2,得到的结果是1、3、9、27……。因此,原始序列的下一项很可能是83。同样,对θ d数据进行简单的变换,或许能帮助我们推测θ d的规律。

At first glance, the numbers seems almost random. Yet subtracting 2 from each term produces 1, 3, 9, 27, . . . Thus, in the original series the next term is likely to be 83. Similarly, a simple transformation of the θd data might help us conjecture a pattern for θd.

图像 θ d数据 的哪种变换会产生简单的模式?

What transformation of the θd data produces simple patterns?

所需的变换应该产生简单的图案,并具有美学或逻辑上的合理性。一个合理性是诚实计算键角的结构,它可以作为两个 C–H 向量的点积来计算(问题 6.3)。由于点积涉及余弦,因此 θ d的一个合理变换是 cos θ d

The desired transformation should produce simple patterns and have aesthetic or logical justification. One justification is the structure of an honest calculation of the bond angle, which can be computed as a dot product of two C–H vectors (Problem 6.3). Because dot products involve cosines, a worthwhile transformation of θd is cos θd.

这种变换简化了数据:cos θ d系列简单地以 −1、−1/2、... 开始。两个可能的延续是 −1/4 或 −1/3;它们分别对应于一般项 −1/2 d−1或 −1/d。

This transformation simplifies the data: The cos θd series begins simply −1, −1/2, . . . Two plausible continuations are −1/4 or −1/3; they correspond, respectively, to the general term −1/2d−1 or −1/d.

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图像 哪种延续和猜想更合理?

Which continuation and conjecture is the more plausible?

两个猜想都预测 cosθ < 0,因此 θd > 90°(对于所有 d)。这个共同的预测令人鼓舞(问题 6.1);然而,共同的预测意味着它无法区分这两个猜想。

Both conjectures predict cos θ < 0 and therefore θd > 90° (for all d). This shared prediction is encouraging (Problem 6.1); however, being shared means that it does not distinguish between the conjectures.

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这两种猜想是否符合分子几何构型?除了键角之外,一个重要的几何特征是碳的位置。在一个维度上,它位于中间两个氢原子之间,因此它将 H–H 线段分成长度比为 1:1 的两段。

Does either conjecture match the molecular geometry? An important geometric feature, apart from the bond angle, is the position of the carbon. In one dimension, it lies halfway between the two hydrogens, so it splits the H–H line segment into two pieces having a 1:1 length ratio.

在二维空间中,碳原子位于连接一个氢原子和另外两个氢原子中点的高度上,碳原子将该高度分成两部分,长度比为 1:2。

In two dimensions, the carbon lies on the altitude that connects one hydrogen to the midpoint of the other two hydrogens. The carbon splits the altitude into two pieces having a 1:2 length ratio.

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图像 碳如何分解出与甲烷类似的高度?

How does the carbon split the analogous altitude of methane?

在甲烷中,类似的高度从顶部顶点延伸至底部中心。碳原子位于平均位置,因此也位于四个氢原子的平均高度。由于三个底部氢原子的高度为零,因此四个氢原子的平均高度为h/4,其中h是顶部氢原子的高度。因此,在三维空间中,碳原子将高度分成两部分,长度比为h/4:3h/4,即1:3。因此,在d维空间中,碳原子可能将高度分成两部分,长度比为1:d(问题6.2)。

In methane, the analogous altitude runs from the top vertex to the center of the base. The carbon lies at the mean position and therefore at the mean height of the four hydrogens. Because the three base hydrogens have zero height, the mean height of the four hydrogens is h/4, where h is the height of the top hydrogen. Thus, in three dimensions, the carbon splits the altitude into two parts having a length ratio of h/4 : 3h/4 or 1 : 3. In d dimensions, therefore, the carbon probably splits the altitude into two parts having a length ratio of 1:d (Problem 6.2).

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由于 1 : d 在几何中自然成立,cos θ d更有可能包含 1/d 而不是 1/2 d−1。因此,两个 cos θ d猜想中,可能性更大的是

Because 1 : d arises naturally in the geometry, cos θd is more likely to contain 1/d rather than 1/2d−1. Thus, the more likely of the two cos θd conjectures is that

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对于甲烷,当 d = 3 时,预测的键角为 arccos(−1/3),约为 109.47°。这个基于类比推理的预测结果与实验结果以及基于解析几何的诚实计算结果(问题 6.3)相符。

For methane, where d = 3, the predicted bond angle is arccos(−1/3) or approximately 109.47°. This prediction using reasoning by analogy agrees with experiment and with an honest calculation using analytic geometry (Problem 6.3).

问题 6.2 碳在高维空间中的位置

Problem 6.2 Carbon’s position in higher dimensions

证明碳将高度分成长度比为 1:d 的两部分的猜想。

Justify conjecture that the carbon splits the altitude into two pieces having a length ratio 1:d.

问题 6.3 解析几何解

Problem 6.3 Analytic-geometry solution

为了用类比法检验解法,请按如下方式运用解析几何来求键角。首先,将坐标 (xn , yn , zn )赋给 n 个氢原子,其中 n = 1 ... 4,然后求解这些坐标。(利用对称性使坐标尽可能简单。)然后选择两个 C–H 向量,计算它们所夹的角。

In order to check the solution using analogy, use analytic geometry as follows to find the bond angle. First, assign coordinates (xn, yn, zn) to the n hydrogens, where n = 1 . . . 4, and solve for those coordinates. (Use symmetry to make the coordinates as simple as you can.) Then choose two C–H vectors and compute the angle that they subtend.

问题 6.4 高维的极端情况

Problem 6.4 Extreme case of high dimensionality

画一幅图来解释小角近似arccos x ≈ π/2 − x。高维空间(d 较大)的近似键角是多少?你能找到近似键角的直观解释吗?

Draw a picture to explain the small-angle approximation arccos x ≈ π/2 − x. What is the approximate bond angle in high dimensions (large d)? Can you find an intuitive explanation for the approximate bond angle?

6.2 拓扑:有多少个区域?

6.2 Topology: How many regions?

甲烷中的键角(第 6.1 节)可以直接用解析几何计算(问题 6.3),因此类比推理并不能充分发挥其威力。因此,请尝试以下问题。

The bond angle in methane (Section 6.1) can be calculated directly with analytic geometry (Problem 6.3), so reasoning by analogy does not show its full power. Therefore, try the following problem.

图像 五个平面将空间划分为多少个区域?

Into how many regions do five planes divide space?

此公式允许退化排列,例如五个平行平面、四个平面交于一点或三个平面交于一条线。为了消除这些及其他退化情况,我们随机放置和调整平面的方向,从而最大化区域数量。问题在于求出五个平面所能生成的区域的最大数量。

This formulation permits degenerate arrangements such as five parallel planes, four planes meeting at a point, or three planes meeting at a line. To eliminate these and other degeneracies, let’s place and orient the planes randomly, thereby maximizing the number of regions. The problem is then to find the maximum number of regions produced by five planes.

五个平面很难想象,但简单情况(使用更少的平面)的方法或许能得出一个推广到五个平面的模式。最简单的情况是零个平面:空间保持完整,因此 R(0) = 1(其中 R(n) 表示 n 个平面产生的区域数)。第一个平面将空间分成两半,得出 R(1) = 2。要添加第二个平面,想象一下将橙子切成两片,得到四个楔形:R(2) = 4。

Five planes are hard to imagine, but the method of easy cases—using fewer planes—might produce a pattern that generalizes to five planes. The easiest case is zero planes: Space remains whole so R(0) = 1 (where R(n) denotes the number of regions produced by n planes). The first plane divides space into two halves, giving R(1) = 2. To add the second plane, imagine slicing an orange twice to produce four wedges: R(2) = 4.

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图像 数据中出现了什么模式?

What pattern(s) appear in the data?

一个合理的推测是 R(n) = 2n 为了验证这一推测,我们可以尝试 n = 3 的情况,将橙子切三遍,将四块橙子分别切成两小块;这样,R(3) 确实等于 8。或许,同样的规律在 R(4) = 16 和 R(5) = 32 时仍然成立。在下表中,这两个外推结果被标记为灰色,以区别于已验证的数值。

A reasonable conjecture is that R(n) = 2n. To test it, try the case n = 3 by slicing the orange a third time and cutting each of the four pieces into two smaller pieces; thus, R(3) is indeed 8. Perhaps the pattern continues with R(4) = 16 and R(5) = 32. In the following table for R(n), these two extrapolations are marked in gray to distinguish them from the verified entries.

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图像 如何进一步检验R(n) = 2 n猜想?

How can the R(n) = 2n conjecture be tested further?

通过计数区域进行直接检验很困难,因为这些区域在三维空间中难以可视化。一个类似的二维问题更容易解决,其解法或许有助于检验三维猜想。二维空间由线分割,因此类似的问题如下:

A direct test by counting regions is difficult because the regions are hard to visualize in three dimensions. An analogous two-dimensional problem would be easier to solve, and its solution may help test the three-dimensional conjecture. A two-dimensional space is partitioned by lines, so the analogous question is the following:

图像 n条线将平面最多划分成多少个区域?

What is the maximum number of regions into which n lines divide the plane?

简单案例法或许能揭示出一种模式。如果模式是 2 n,那么 R(n) = 2 n猜想很可能在三维空间中适用。

The method of easy cases might suggest a pattern. If the pattern is 2n, then the R(n) = 2n conjecture is likely to apply in three dimensions.

图像 在一些简单的情况下会发生什么?

What happens in a few easy cases?

零线离开平面完整,给出 R(0) = 1。接下来的三种情况如下(尽管参见问题 6.5):

Zero lines leave the plane whole, giving R(0) = 1. The next three cases are as follows (although see Problem 6.5):

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问题 6.5 又是三行

Problem 6.5 Three lines again

R(3) = 7 的图示表明三条线产生了七个区域。这是另一个包含三条线的例子,也是随机排列的,但似乎只产生了六个区域。那么第七个区域在哪里呢?或者 R(3) = 6 呢?

The R(3) = 7 illustration showed three lines producing seven regions. Here is another example with three lines, also in a random arrangement, but it seems to produce only six regions. Where, if anywhere, is the seventh region? Or is R(3) = 6?

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问题 6.6 凸性

Problem 6.6 Convexity

所有由线段创建的区域都必须是凸的吗?(当且仅当连接区域内任意两点的线段完全位于区域内时,该区域才是凸的。)那么通过在空间中放置平面创建的三维区域呢?

Must all the regions created by the lines be convex? (A region is convex if and only if a line segment connecting any two points inside the region lies entirely inside the region.) What about the three-dimensional regions created by placing planes in space?

在 R(3) 被证明为 7 之前,R(n) = 2 n猜想看起来是成立的。然而,在放弃这样一个简单的猜想之前,画一条第四条线,仔细数一数其中的区域。四条线只构成了 11 个区域,而不是预测的 16 个,因此 2 n猜想已经失效。

Until R(3) turned out to be 7, the conjecture R(n) = 2n looked sound. However, before discarding such a simple conjecture, draw a fourth line and carefully count the regions. Four lines make only 11 regions rather than the predicted 16, so the 2n conjecture is dead.

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当看到二维数据 R 2 (n) 和三维数据 R 3 (n) 时,可能会产生新的猜想。

A new conjecture might arise from seeing the two-dimensional data R2(n) alongside the three-dimensional data R3(n).

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在此表中,多个条目组合起来形成相近的条目。例如,R 2 (1) 和 R 3 (1)(n = 1 列中的两个条目)加起来等于 R 2 (2) 或 R 3 (2)。这两个条目又加起来等于 R 3 (3)。但是,该表中有许多小数,并且有多种组合方式;要排除这些巧合,需要收集更多数据——最简单的数据源就是类似的一维问题。

In this table, several entries combine to make nearby entries. For example, R2(1) and R3(1)—the two entries in the n = 1 column—sum to R2(2) or R3(2). These two entries in turn sum to the R3(3) entry. But the table has many small numbers with many ways to combine them; discarding the coincidences requires gathering further data—and the simplest data source is the analogous one-dimensional problem.

图像 n 个点将一条线最多分成多少个线段

What is the maximum number of segments into which n points divide a line?

一个诱人的答案是,n 个点组成 n 条线段。然而,一个简单的情况——一个点组成两条线段——就降低了这个答案的吸引力。更确切地说,n 个点组成 n + 1 条线段。这个结果生成了下表中的R 1行。

A tempting answer is that n points make n segments. However, an easy case—that one point produces two segments—reduces the temptation. Rather, n points make n + 1 segments. That result generates the R1 row in the following table.

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图像 这些数据中存在什么模式?

What patterns are in these data?

2n猜想部分成立。在 R1中,它从 n = 2 开始失效。在 R2中,它从 n = 3 开始失效。因此,在 R3 行中很可能从 n = 4 开始失效,这使得 R3(4) = 16 和 R3(5) = 32 这两个猜想不太可能成立。我个人估计,在发现这些猜想失效之前,R3(4) = 16 猜想的概率为0.5;但现在它最多降至 0.01。(更多关于估计和更新猜想概率的信息,请参阅 Corfield [ 11 ]、Jaynes [ 21 ] 和 Polya [ 36 ]关于合情推理的重要著作。)

The 2n conjecture survives partially. In the R1 row, it fails starting at n = 2. In the R2 row, it fails starting at n = 3. Thus in the R3 row, it probably fails starting at n = 4, making the conjectures R3(4) = 16 and R3(5) = 32 improbable. My personal estimate is that, before seeing these failures, the probability of the R3(4) = 16 conjecture was 0.5; but now it falls to at most 0.01. (For more on estimating and updating the probabilities of conjectures, see the important works on plausible reasoning by Corfield [11], Jaynes [21], and Polya [36].)

更好的消息是,这些明显的巧合蕴含着一个稳健的模式:

In better news, the apparent coincidences contain a robust pattern:

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图像 如果按照这个模式继续下去,五个平面可以将空间划分成多少个区域?

If the pattern continues, into how many regions can five planes divide space?

根据模式,

According to the pattern,

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进而

and then

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因此,五个平面最多可以将空间划分为26个区域。

Thus, five planes can divide space into a maximum of 26 regions.

通过绘制五个平面并计算区域数来推导这个数字非常困难。此外,这种蛮力推导只能得出 R 3 (5) 的值,而简单的情况和类比方法可以计算表中的任意项。因此,它们提供了足够的数据来推测 R 2 (n)(问题 6.9)、R 3 (n)(问题 6.10)以及一般项 R d (n)(问题 6.12)的表达式。

This number is hard to deduce by drawing five planes and counting the regions. Furthermore, that brute-force approach would give the value of only R3(5), whereas easy cases and analogy give a method to compute any entry in the table. They thereby provide enough data to conjecture expressions for R2(n) (Problem 6.9), for R3(n) (Problem 6.10), and for the general entry Rd(n) (Problem 6.12).

问题 6.7 检查二维模式

Problem 6.7 Checking the pattern in two dimensions

推测的模式预测 R² ( 5) = 16:五条线可以将平面分成 16 个区域。通过画五条线并计算区域数量来验证该推测。

The conjectured pattern predicts R2(5) = 16: that five lines can divide the plane into 16 regions. Check the conjecture by drawing five lines and counting the regions.

问题 6.8 从零维中释放数据

Problem 6.8 Free data from zero dimensions

由于一维问题给出了有用的数据,不妨尝试零维问题。将 R 3、R 2和 R 1行的模式向上扩展,构造一个 R 0行。它给出了用 n 个对象(维度为 −1)划分一个点所产生的零维区域(点)的数量。如果该行遵循观察到的模式,那么 R 0是多少?这个结果是否与尝试细分点的几何意义一致?

Because the one-dimensional problem gave useful data, try the zero-dimensional problem. Extend the pattern for the R3, R2, and R1 rows upward to construct an R0 row. It gives the number of zero-dimensional regions (points) produced by partitioning a point with n objects (of dimension −1). What is R0 if the row is to follow the observed pattern? Is that result consistent with the geometric meaning of trying to subdivide a point?

问题 6.9 二维的一般结果

Problem 6.9 General result in two dimensions

R 0数据拟合 R 0 (n) = 1(问题 6.8),这是一个零次多项式。R 1数据拟合 R 1 (n) = n + 1,这是一个一次多项式。因此,R 2数据可能拟合二次多项式。

The R0 data fits R0(n) = 1 (Problem 6.8), which is a zeroth-degree polynomial. The R1 data fits R1(n) = n + 1, which is a first-degree polynomial. Therefore, the R2 data probably fits a quadratic.

通过将 n = 0 ... 2 的数据拟合到一般二次方程 An 2 + Bn + C 来检验这个猜想,反复取出大的部分(第 5 章),如下所示。

Test this conjecture by fitting the data for n = 0 . . . 2 to the general quadratic An2 + Bn + C, repeatedly taking out the big part (Chapter 5) as follows.

a. 猜测二次系数 A 的一个合理值。然后取出(减去)较大的部分 An 2,并将剩余部分 R 2 (n) − An 2制成表格,其中 n = 0 ... 2。如果余项与 n 不是线性关系,则表示存在二次项,或者移除了太多项。无论哪种情况,都需要调整 A。

a. Guess a reasonable value for the quadratic coefficient A. Then take out (subtract) the big part An2 and tabulate the leftover, R2(n) − An2, for n = 0 . . . 2. If the leftover is not linear in n, then a quadratic term remains or too much was removed. In either case, adjust A.

b. 一旦二次系数 A 正确,使用类似的程序来查找线性系数 B。

b. Once the quadratic coefficient A is correct, use an analogous procedure to find the linear coefficient B.

c. 同理求解常数系数C。

c. Similarly solve for the constant coefficient C.

d. 用新数据检查二次拟合(R 2 (n),n ≥ 3)。

d. Check your quadratic fit against new data (R2(n) for n ≥ 3).

问题 6.10 三维中的一般结果

Problem 6.10 General result in three dimensions

一个合理的推测是,R 3行与三次函数匹配(问题 6.9)。使用去掉较大部分的方法,将三次函数拟合到 n = 0 ... 3 的数据。它是否得到了推测值 R 3 (4) = 15 和 R 3 (5) = 26?

A reasonable conjecture is that the R3 row matches a cubic (Problem 6.9). Use taking out the big part to fit a cubic to the n = 0 . . . 3 data. Does it produce the conjectured values R3(4) = 15 and R3(5) = 26?

问题 6.11 几何解释

Problem 6.11 Geometric explanation

找出观察到的模式的几何解释。提示:首先解释为什么该模式会从 R 1行生成 R 2行;然后推广该原因来解释 R 3行。

Find a geometric explanation for the observed pattern. Hint: Explain first why the pattern generates the R2 row from the R1 row; then generalize the reason to explain the R3 row.

问题 6.12 任意维的通解

Problem 6.12 General solution in arbitrary dimension

连接 R d (n) 表中相邻条目的模式是生成帕斯卡三角形的模式 [ 17 ]。由于帕斯卡三角形产生二项式系数,因此一般表达式 R d (n) 应该包含二项式系数。

The pattern connecting neighboring entries of the Rd(n) table is the pattern that generates Pascal’s triangle [17]. Because Pascal’s triangle produces binomial coefficients, the general expression Rd(n) should contain binomial coefficients.

因此,用二项式系数表示R 0 (n)(问题6.8)、R 1 (n)和R 2 (n)(问题6.9 )。然后推测R 3 (n)和R d (n)的二项式系数形式,并对照问题6.10检查结果。

Therefore, use binomial coefficients to express R0(n) (Problem 6.8), R1(n), and R2(n) (Problem 6.9). Then conjecture a binomial-coefficient form for R3(n) and Rd(n), checking the result against Problem 6.10.

问题 6.13 2 的幂猜想

Problem 6.13 Power-of-2 conjecture

我们对区域数量的第一个猜想是 R d (n) = 2 n。在三维空间中,直到 n = 4 时,该猜想成立。在 d 维空间中,证明当 n ≤ d 时,R d (n) = 2 n (或许可以使用问题 6.12的结果)。

Our first conjecture for the number of regions was Rd(n) = 2n. In three dimensions, it worked until n = 4. In d dimensions, show that Rd(n) = 2n for n ≤ d (perhaps using the results of Problem 6.12).

6.3 算子:欧拉-麦克劳林求和

6.3 Operators: Euler–MacLaurin summation

下一个类比研究的是一些不常见的函数。大多数函数会将数字转换为其他数字,但有些特殊类型的函数——运算符——会将函数转换为其他函数。一个常见的例子是导数运算符 D。它将正弦函数转换为余弦函数,或将双曲正弦函数转换为双曲余弦函数。在运算符符号中,D(sin) = cos 和 D(sin h) = cosh;省略括号可以得到更简洁的表达式 D sin = cos 和 D sin h = cosh。要理解和学习如何使用运算符,一个有效的工具是类比推理:运算符的行为与普通函数非常相似,甚至与数字也类似。

The next analogy studies unusual functions. Most functions turn numbers into other numbers, but special kinds of functions—operators—turn functions into other functions. A familiar example is the derivative operator D. It turns the sine function into the cosine function, or the hyperbolic sine function into the hyperbolic cosine function. In operator notation, D(sin) = cos and D(sin h) = cosh; omitting the parentheses gives the less cluttered expression D sin = cos and D sin h = cosh. To understand and learn how to use operators, a fruitful tool is reasoning by analogy: Operators behave much like ordinary functions or even like numbers.

6.3.1 左移

6.3.1 Left shift

与数字类似,导数算子 D 可以平方得到 D 2(二阶导数算子),也可以得到 D 的任意整数幂。同样,导数算子也可以用于多项式。在这种情况下,一个普通的多项式,例如 P(x) = x 2 + x/10 + 1,可以得到算子多项式 P(D) = D 2 + D/10 + 1(小阻尼弹簧质量系统的微分算子)。

Like a number, the derivative operator D can be squared to make D2 (the second-derivative operator) or to make any integer power of D. Similarly, the derivative operator can be fed to a polynomial. In that usage, an ordinary polynomial such as P(x) = x2 + x/10 + 1 produces the operator polynomial P(D) = D2 + D/10 + 1 (the differential operator for a lightly damped spring–mass system).

与数字的类比能延伸到什么程度?例如,cosh D 或 sin D 有意义吗?因为这些函数可以用指数函数写成,所以我们来研究一下指数 e D运算符。

How far does the analogy to numbers extend? For example, do cosh D or sin D have a meaning? Because these functions can be written using the exponential function, let’s investigate the operator exponential eD.

图像 e D 是什么意思?

What does eD mean?

e D的直接解释是,它将函数 f 转换为 e Df

The direct interpretation of eD is that it turns a function f into eDf.

图像

然而,这种解释毫无必要地是非线性的。它将 2f 转化为 e 2Df ,即 e Df的平方,而一个由 f 生成 e Df 的线性算子,会由 2f 生成 2e Df。为了得到线性解释,可以使用泰勒级数(假设 D 是一个数字)来从线性算子中构造 e D 。

However, this interpretation is needlessly nonlinear. It turns 2f into e2Df, which is the square of eDf, whereas a linear operator that produces eDf from f would produce 2eDf from 2f. To get a linear interpretation, use a Taylor series—as if D were a number—to build eD out of linear operators.

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图像 这个e D 对简单函数有什么作用?

What does this eD do to simple functions?

最简单的非零函数是常数函数 f = 1。这是将该函数输入到 e D中:

The simplest nonzero function is the constant function f = 1. Here is that function being fed to eD:

图像

下一个最简单的函数 x 变成 x + 1。

The next simplest function x turns into x + 1.

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更有趣的是,x 2变成 (x + 1) 2

More interestingly, x2 turns into (x + 1)2.

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问题 6.14 继续该模式

Problem 6.14 Continue the pattern

e D x 3是多少?一般来说,e D x n是多少?

What is eDx3 and, in general, eDxn?

图像 e D 通常做什么?

What does eD do in general?

前面的例子遵循 e D x n = (x+1) n的模式。由于 x 的大多数函数都可以展开为 x 的幂,并且 e D将 x n 的每一项都转换为 (x + 1) n,因此结论是 e D将 f(x) 转换为 f(x + 1)。令人惊讶的是,e D就是 L,即左移运算符。

The preceding examples follow the pattern eDxn = (x+1)n. Because most functions of x can be expanded in powers of x, and eD turns each xn term into (x + 1)n, the conclusion is that eD turns f(x) into f(x + 1). Amazingly, eD is simply L, the left-shift operator.

问题 6.15 右移或左移

Problem 6.15 Right or left shift

画图表明 f(x) → f(x + 1) 是左移而非右移。将 e −D应用于几个简单函数来描述其行为。

Draw a graph to show that f(x) → f(x + 1) is a left rather than a right shift. Apply e−D to a few simple functions to characterize its behavior.

问题 6.16 对更难的函数进行操作

Problem 6.16 Operating on a harder function

将 e D的泰勒展开式应用于sin x,得出 e D sin x = sin(x + 1)。

Apply the Taylor expansion for eD to sin x to show that eD sin x = sin(x + 1).

问题 6.17 通用移位运算符

Problem 6.17 General shift operator

如果 x 有维度,则导数算子 D = d/dx 不是无维度的,并且 e D是非法表达式。为了使通式 e aD合法,a 的维度必须是多少?e aD的作用是什么?

If x has dimensions, then the derivative operator D = d/dx is not dimensionless, and eD is an illegal expression. To make the general expression eaD legal, what must the dimensions of a be? What does eaD do?

6.3.2 总结

6.3.2 Summation

正如导数算子可以表示左移算子(如 L = e D),左移算子也可以表示求和运算。这种算子表示将带来一种强大的方法,用于近似没有封闭形式的和。

Just as the derivative operator can represent the left-shift operator (as L = eD), the left-shift operator can represent the operation of summation. This operator representation will lead to a powerful method for approximating sums with no closed form.

求和类似于我们更熟悉的积分运算。积分有定积分和不定积分两种形式:定积分等价于不定积分,然后在积分极限处求值。例如,这是 f(x) = 2x 的定积分。

Summation is analogous to the more familiar operation of integration. Integration occurs in definite and indefinite flavors: Definite integration is equivalent to indefinite integration followed by evaluation at the limits of integration. As an example, here is the definite integration of f(x) = 2x.

图像

一般来说,输入函数 g 与不定积分结果之间的关系是 DG = g,其中 D 是导数算子,G = ∫g 是不定积分的结果。因此,D 和 ∫ 互为逆函数彼此之间——D ∫ = 1 或 D = 1/∫——图中的环路表示这种联系。(由于可能存在积分常数,∫D ≠ 1。)

In general, the connection between an input function g and the result of indefinite integration is DG = g, where D is the derivative operator and G = ∫g is the result of indefinite integration. Thus D and ∫ are inverses of one another—D ∫ = 1 or D = 1/∫—a connection represented by the loop in the diagram. (∫D ≠ 1 because of a possible integration constant.)

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图像 总结的类似图像是什么?

What is the analogous picture for summation?

与积分类似,将定积分定义为不定积分,然后在极限处求值。但应用类比时要谨慎,以免出现偏差一或栅栏柱误差(问题 2.24)。该和图像包含三个矩形——f(2)、f(3) 和 f(4),而定积分图像dk 不包含 f(4) 的任何矩形。与其通过重新定义我们熟悉的积分运算来纠正这种差异,不如将不定积分解释为排除最后一个矩形。然后,进行不定积分,并在极限 a 和 b 处求值,得到的和的指标范围从 a 到 b − 1。

Analogously to integration, define definite summation as indefinite summation and then evaluation at the limits. But apply the analogy with care to avoid an off-by-one or fencepost error (Problem 2.24). The sum includes three rectangles—f(2), f(3), and f(4)—whereas the definite integral dk does not include any of the f(4) rectangle. Rather than rectifying the discrepancy by redefining the familiar operation of integration, interpret indefinite summation to exclude the last rectangle. Then indefinite summation followed by evaluating at the limits a and b produces a sum whose index ranges from a to b − 1.

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举个例子,取 f(k) = k。那么不定和 Σf 就是函数 F,其定义为 F(k) = k(k−1)/2+C(其中 C 是求和常数)。在 0 和 n 之间求 F 的值可得出 n(n − 1)/2,即图像。在下图中,这些步骤是正向路径。

As an example, take f(k) = k. Then the indefinite sum Σf is the function F defined by F(k) = k(k−1)/2+C (where C is the constant of summation). Evaluating F between 0 and n gives n(n − 1)/2, which is . In the following diagram, these steps are the forward path.

图像

在逆向路径中,新的 Δ 算子使 Σ 反转,就像微分使积分反转一样。因此,Δ 的算子表示形式也为 Σ 提供了表示形式。由于 Δ 和导数算子 D 类似,因此它们的表示形式可能也类似。导数是极限

In the reverse path, the new Δ operator inverts Σ just as differentiation inverts integration. Therefore, an operator representation for Δ provides one for Σ. Because Δ and the derivative operator D are analogous, their representations are probably analogous. A derivative is the limit

图像

因此,导数算子 D 是算子极限

The derivative operator D is therefore the operator limit

图像

其中 L h运算符将 f(x) 转换为 f(x + h),即 L h左移 h。

where the Lh operator turns f(x) into f(x + h)—that is, Lh left shifts by h.

问题 6.18 运算符限制

Problem 6.18 Operator limit

解释为什么当 h 较小时,L h ≈ 1 + hD。由此证明 L = e D

Explain why Lh ≈ 1 + hD for small h. Show therefore that L = eD.

图像 Δ 的类似表示是什么?

What is an analogous representation of Δ?

D 的算子极限使用无穷小左移;相应地,积分的逆运算是对无穷小宽度的矩形求和。由于求和 Σ 对单位宽度的矩形求和,其逆 Δ 应该使用单位左移,即 L h,其中 h = 1。合理的推测是:

The operator limit for D uses an infinitesimal left shift; correspondingly, the inverse operation of integration sums rectangles of infinitesimal width. Because summation Σ sums rectangles of unit width, its inverse Δ should use a unit left shift—namely, Lh with h = 1. As a reasonable conjecture,

图像

这个 Δ(称为有限差分算子)被构造为 1/Σ。如果构造正确,则 (L − 1)Σ 是恒等算子 1。换句话说,(L − 1)Σ 应该将函数转化为自身。

This Δ—called the finite-difference operator—is constructed to be 1/Σ. If the construction is correct, then (L − 1)Σ is the identity operator 1. In other words, (L − 1)Σ should turn functions into themselves.

图像 这个猜想在各种简单情况下效果如何?

How well does this conjecture work in various easy cases?

为了检验这个猜想,首先将算子 (L−1)Σ 应用于简单函数 g = 1。然后 Σg 是一个等待输入参数的函数,而 (Σg)(k) 是输入 k 的结果。按照这个符号,(Σg)(k) = k + C。将此函数输入 L − 1 算子可以重现 g。

To test the conjecture, apply the operator (L−1)Σ first to the easy function g = 1. Then Σg is a function waiting to be fed an argument, and (Σg)(k) is the result of feeding it k. With that notation, (Σg)(k) = k + C. Feeding this function to the L − 1 operator reproduces g.

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使用次简单的函数(定义为 g(k) = k),不定和 (Σg)(k) 为 k(k − 1)/2 + C。将 Σg 再次经过 L − 1 可重现 g。

With the next-easiest function—defined by g(k) = k—the indefinite sum (Σg)(k) is k(k − 1)/2 + C. Passing Σg through L − 1 again reproduces g.

图像

综上所述,对于测试函数 g(k) = 1 和 g(k) = k,算子乘积 (L − 1)Σ 将 g 取回自身,因此其作用类似于恒等算子。

In summary, for the test functions g(k) = 1 and g(k) = k, the operator product (L − 1)Σ takes g back to itself, so it acts like the identity operator.

这种行为具有普遍性——(L−1)Σ1 确实为 1,且 Σ= 1/(L−1)。由于 L = e D,因此有 Σ= 1/(e D − 1)。将泰勒级数的右侧展开,可以得到求和算符的惊人表示。

This behavior is general—(L−1)Σ1 is indeed 1, and Σ= 1/(L−1). Because L = eD, we have Σ= 1/(eD − 1). Expanding the right side in a Taylor series gives an amazing representation of the summation operator.

图像

因为 D ∫ = 1,首项 1/D 是积分。因此,求和近似于积分——这是一个合理的结论,表明算子表示并非无稽之谈。

Because D ∫ = 1, the leading term 1/D is integration. Thus, summation is approximately integration—a plausible conclusion indicating that the operator representation is not nonsense.

将此算子系列应用于函数 f,然后在极限 a 和 b 处求值,得到欧拉-麦克劳林求和公式

Applying this operator series to a function f and then evaluating at the limits a and b produces the Euler–MacLaurin summation formula

图像

其中 f (n)表示 f 的 n 阶导数。

where f(n) indicates the nth derivative of f.

和式中缺少通常的最后一个项 f(b)。包含这一项,可以得到一个有用的替代式

The sum lacks the usual final term f(b). Including this term gives the useful alternative

图像

为了检验,尝试一个简单的例子:图像。使用欧拉-麦克劳林求和法,f(k) = k,a = 0,b = n。积分项贡献 n 2 /2;常数项 [f(b) + f(a)]/2 贡献 n/2;其余项为零。结果熟悉且正确:

As a check, try an easy case: . Using Euler–MacLaurin summation, f(k) = k, a = 0, and b = n. The integral term then contributes n2/2; the constant term [f(b) + f(a)]/2 contributes n/2; and later terms vanish. The result is familiar and correct:

图像

对欧拉-麦克劳林求和法更严格的检验是近似ln n!,即和图像k(见第4.5节)。因此,在(含)a = 1和b = n之间的和f(k) = ln k。结果为

A more stringent test of Euler–MacLaurin summation is to approximate ln n!, which is the sum k (Section 4.5). Therefore, sum f(k) = ln k between the (inclusive) limits a = 1 and b = n. The result is

图像

1/D算子的积分贡献了ln k曲线下的面积。1/2算子的修正包含了三角形凸起(问题6.20)。省略号包含高阶修正(问题6.21)——用图片很难计算(问题4.32),但用欧拉-麦克劳林求和(问题6.21)很容易计算。

The integral, from the 1/D operator, contributes the area under the ln k curve. The correction, from the 1/2 operator, incorporates the triangular protrusions (Problem 6.20). The ellipsis includes the higher-order corrections (Problem 6.21)—hard to evaluate using pictures (Problem 4.32) but simple using Euler–MacLaurin summation (Problem 6.21).

图像

问题 6.19 整数和

Problem 6.19 Integer sums

使用欧拉-麦克劳林求和法找到以下和的闭合形式:

Use Euler–MacLaurin summation to find closed forms for the following sums:

图像

问题 6.20 边界情况

Problem 6.20 Boundary cases

在欧拉-麦克劳林求和法中,常数项是 [f(b) + f(a)]/2——第一项的一半加上最后一项的一半。ln k 求和图(4.5 节)显示,凸起部分大约是最后一项的一半,即 ln n。从图中可以看出,第一项的一半发生了什么?

In Euler–MacLaurin summation, the constant term is [f(b) + f(a)]/2—one-half of the first term plus one-half of the last term. The picture for summing ln k (Section 4.5) showed that the protrusions are approximately one-half of the last term, namely ln n. What, pictorially, happened to one-half of the first term?

问题 6.21 高阶项

Problem 6.21 Higher-order terms

使用欧拉-麦克劳林求和法近似 ln 5!。

Approximate ln 5! using Euler–MacLaurin summation.

问题 6.22 巴塞尔总和

Problem 6.22 Basel sum

巴塞尔总和图像可以用图片来近似表示(问题 4.37)。

The Basel sum may be approximated with pictures (Problem 4.37).

然而,这种近似过于粗糙,无法帮助猜测其闭合形式。像欧拉那样,使用欧拉-麦克劳林求和来提高精度,直到你能自信地猜出闭合形式。提示:明确地求出前几项的和。

However, the approximation is too crude to help guess the closed form. As Euler did, use Euler–MacLaurin summation to improve the accuracy until you can confidently guess the closed form. Hint: Sum the first few terms explicitly.

6.4 正切根:一个令人畏惧的超越和

6.4 Tangent roots: A daunting transcendental sum

我们之所以选择这个告别的例子,是因为它的分析结合了多种街头斗争工具,它是一个难以解决的无限总和。

Our farewell example, chosen because its analysis combines diverse street-fighting tools, is a difficult infinite sum.

找出图像x n是 tan x = x 的正解。

Find where the xn are the positive solutions of tan x = x.

tan x = x 的解,或者等价地,tan x − x 的根,是超越函数,没有封闭形式,然而几乎每种求和法都需要封闭形式。街头斗争法会来帮我们。

The solutions to tan x = x or, equivalently, the roots of tan x − x, are transcendental and have no closed form, yet a closed form is required for almost every summation method. Street-fighting methods will come to our rescue.

6.4.1 图片和简单案例

6.4.1 Pictures and easy cases

从一个简单的案例开始分析。

Begin the analysis with a hopefully easy case.

图像 x 1 的第一个根是多少

What is the first root x1?

tan x − x 的根由 y = x 与 y = tan x 的交点给出。令人惊讶的是,tan x 的分支中没有交点,当 0 < x < π/2 时(习题 6.23),第一个交点恰好位于 x = 3π/2 处的渐近线之前。因此,x 1 ≈ 3π/2。

The roots of tan x − x are given by the intersections of y = x and y = tan x. Surprisingly, no intersection occurs in the branch of tan x where 0 < x < π/2 (Problem 6.23); the first intersection is just before the asymptote at x = 3π/2. Thus, x1 ≈ 3π/2.

图像

问题 6.23 与主分支无交点

Problem 6.23 No intersection with the main branch

用符号证明,tan x = x 在 0 < x < π/2 处无解。(该结果在图中看起来合理,但为了绘制图形,值得检查一下。)

Show symbolically that tan x = x has no solution for 0 < x < π/2. (The result looks plausible pictorially but is worth checking in order to draw the picture.)

图像 后续交叉路口大概在哪里?

Where, approximately, are the subsequent intersections?

随着 x 的增长,y = x 直线与 y = tan x 图像的交点越来越高,因此也越来越接近垂直渐近线。因此,对 x n的较大部分进行以下渐近线近似:

As x grows, the y = x line intersects the y = tan x graph ever higher and therefore ever closer to the vertical asymptotes. Therefore, make the following asymptote approximation for the big part of xn:

图像

6.4.2 取出大部件

6.4.2 Taking out the big part

x n的这个近似的低熵表达式给出了 S 的大部分(零近似值)。

This approximate, low-entropy expression for xn gives the big part of S (the zeroth approximation).

图像

图像从图片(第 4.5 节)或欧拉-麦克劳林求和(第 6.3.2 节)可以看出,该总和大致为以下积分。

The sum is, from a picture (Section 4.5) or from Euler–MacLaurin summation (Section 6.3.2), roughly the following integral.

图像

所以,

Therefore,

图片

阴影部分的凸起大致是三角形,它们的总面积是第一个矩形的一半。该矩形的面积为 1/9,所以

The shaded protrusions are roughly triangles, and they sum to one-half of the first rectangle. That rectangle has area 1/9, so

图片

图片

因此,S 的更准确估计是

Therefore, a more accurate estimate of S is

图片

这比最初的估计略高。

which is slightly higher than the first estimate.

图像 新的近似值是高估了还是低估了?

Is the new approximation an overestimate or an underestimate?

新的近似基于两个低估。首先,渐近线近似 x n ≈ (n + 0.5)π 高估了每个 x n,因此低估了和式中的倒数平方图片;其次,在进行渐近线近似后,和式的图形近似图片用内接三角形代替了每个凸起,从而低估了每个凸起(问题 6.24)。

The new approximation is based on two underestimates. First, the asymptote approximation xn ≈ (n + 0.5)π overestimates each xn and therefore underestimates the squared reciprocals in the sum Second, after making the asymptote approximation, the pictorial approximation to the sum replaces each protrusion with an inscribed triangle and thereby underestimates each protrusion (Problem 6.24).

问题 6.24 第二次低估的图片

Problem 6.24 Picture for the second underestimate

画出图形近似中低估的部分

Draw a picture of the underestimate in the pictorial approximation

图片

图像 如何弥补这两个低估?

How can these two underestimates be remedied?

第二个低估(凸起部分)可以通过图片精确求和来消除。这个和式之所以不为人熟知,部分原因在于它的第一项是分数 1/9——它的任意性增加了和式的熵。加上 n = 0 项(即 1)和偶数平方倒数 1/(2n) ²,就得到了一个简洁而熟悉的低熵和式。

The second underestimate (the protrusions) is eliminated by summing exactly. The sum is unfamiliar partly because its first term is the fraction 1/9—whose arbitrariness increases the entropy of the sum. Including the n = 0 term, which is 1, and the even squared reciprocals 1/(2n)2 produces a compact and familiar lower-entropy sum.

图片

最终的低熵和就是著名的巴塞尔和(高熵结果通常并不出名)。它的值为 B = π 2 /6(问题 6.22)。

The final, low-entropy sum is the famous Basel sum (high-entropy results are not often famous). Its value is B = π2/6 (Problem 6.22).

图像 知道B = π 2 /6如何帮助计算原始和 图片

How does knowing B = π2/6 help evaluate the original sum ?

对原始和的主要修改是加入了偶数平方倒数。它们的和是 B/4。

The major modification from the original sum was to include the even squared reciprocals. Their sum is B/4.

图片

第二个修改是加入n = 0项。因此,为了得到,需要调整Basel值B,方法是先减去B/4,然后再减去n = 0项。代入B = π 2 /6图片后,结果为

The second modification was to include the n = 0 term. Thus, to obtain , adjust the Basel value B by subtracting B/4 and then the n = 0 term. The result, after substituting B = π2/6, is

图片

基于 x n的渐近线近似,此精确总和可得出以下 S 估计值。

This exact sum, based on the asymptote approximation for xn, produces the following estimate of S.

图片

通过扩展产品进行简化,得到

Simplifying by expanding the product gives

图片

问题 6.25 检查之前的推理

Problem 6.25 Check the earlier reasoning

检查前面的图形推理(问题 6.24),1/6 + 1/18 = 2/9 低估了图片该估计的准确度如何?

Check the earlier pictorial reasoning (Problem 6.24) that 1/6 + 1/18 = 2/9 underestimates How accurate was that estimate?

这是使用渐近线近似 x n ≈ (n + 0.5)π 的第三个 S 估计值。综合起来,这些估计值如下

This estimate of S is the third that uses the asymptote approximation xn ≈ (n + 0.5)π. Assembled together, the estimates are

图片

因为第三个估计值包含了的精确值图片,所以 S 估计值中的任何剩余误差都必须属于渐近线近似值本身。

Because the third estimate incorporated the exact value of , any remaining error in the estimate of S must belong to the asymptote approximation itself.

图像 对于哪一项, 图片 渐近线近似最不准确?

For which term of is the asymptote approximation most inaccurate?

随着 x 的增长,x 和 tan x 的图形交点越来越接近垂直渐近线。因此,渐近线近似在 n = 1 时产生最大的绝对误差。由于 x 1是最小根,因此 x n的分数误差相对于 x n的绝对误差,在 n = 1 时更加集中。 的分数误差是 x n图片分数误差的 −2 倍(见 5.3 节),在 n = 1 时同样集中。因为(分数误差乘以自身)在 n = 1 时是最大的。图片图片

As x grows, the graphs of x and tan x intersect ever closer to the vertical asymptote. Thus, the asymptote approximation makes its largest absolute error when n = 1. Because x1 is the smallest root, the fractional error in xn is, relative to the absolute error in xn, even more concentrated at n = 1. The fractional error in , being −2 times the fractional error in xn (Section 5.3), is equally concentrated at n = 1. Because is the (the fractional error times itself) is, by far, the largest at n = 1.

问题 6.26 早期项中的绝对误差

Problem 6.26 Absolute error in the early terms

图片估计渐近线近似产生的绝对误差,作为 n 的函数。

Estimate, as a function of n, the absolute error in that is produced by the asymptote approximation.

由于误差集中在 n = 1 处,S 估计值的最大改进来自于用更精确的值替换近似值 x 1 = (n + 0.5)π。一种简单的数值方法是使用牛顿-拉夫森法进行逐次逼近(问题 4.38)。要用这种方法求根,首先对 x 进行一个初始猜测,然后使用替换值反复改进它

With the error so concentrated at n = 1, the greatest improvement in the estimate of S comes from replacing the approximation x1 = (n + 0.5)π with a more accurate value. A simple numerical approach is successive approximation using the Newton–Raphson method (Problem 4.38). To find a root with this method, make a starting guess x and repeatedly improve it using the replacement

图片

当 x 的起始猜测值略低于 1.5π 处的第一个渐近线时,该过程迅速收敛到 x 1 = 4.4934...

When the starting guess for x is slightly below the first asymptote at 1.5π, the procedure rapidly converges to x1 = 4.4934 . . .

因此,为了改进基于渐近线近似的估计值 S ≈ 0.094715,减去其近似的第一项(其较大部分)并添加校正后的第一项。

Therefore, to improve the estimate S ≈ 0.094715, which was based on the asymptote approximation, subtract its approximate first term (its big part) and add the corrected first term.

图片

此外,使用牛顿-拉夫森法进行改进,1/x 2 2项得出 S ≈ 0.09978(问题 6.27)。因此,一个合理的猜测是

Using the Newton–Raphson method to refine, in addition, the 1/x22 term gives S ≈ 0.09978 (Problem 6.27). Therefore, a highly educated guess is

图片

未知超越数的无穷和似乎既不是超越数,也不是无理数!这个简单而令人惊讶的有理数值得一个简单的解释。

The infinite sum of unknown transcendental numbers seems to be neither transcendental nor irrational! This simple and surprising rational number deserves a simple explanation.

问题 6.27 继续修正

Problem 6.27 Continuing the corrections

选择一个小的 N,比如 4。然后使用牛顿-拉夫逊法计算 n = 1 ... N 时 x n的精确值;并使用这些值来改进 S 的估计值。当你将计算扩展到更大的 N 值时,改进后的 S 估计值是否接近我们的有根据的猜测 1/10?

Choose a small N, say 4. Then use the Newton-Raphson method to compute accurate values of xn for n = 1 . . . N; and use those values to refine the estimate of S. As you extend the computation to larger values of N, do the refined estimates of S approach our educated guess of 1/10?

6.4.3 与多项式的类比

6.4.3 Analogy with polynomials

如果方程 tan x − x = 0 只有几个闭式解就好了!这样,和 S 就很容易计算了。用一个有单根的多项式方程代替 tan x − x 就可以实现这个愿望。最简单的有趣多项式是二次多项式,所以可以用一个简单的二次方程来做实验,例如 x 2 − 3x + 2。

If only the equation tan x − x = 0 had just a few closed-form solutions! Then the sum S would be easy to compute. That wish is fulfilled by replacing tan x − x with a polynomial equation with simple roots. The simplest interesting polynomial is the quadratic, so experiment with a simple quadratic—for example, x2 − 3x + 2.

该多项式有两个根,x 1 = 1 和 x 2 = 2;因此图片,与正切根和类似的多项式根和有两个项。

This polynomial has two roots, x1 = 1 and x2 = 2; therefore , the polynomial-root sum analog of the tangent-root sum, has two terms.

图片

这种计算根和的蛮力方法需要二次方程的解。然而,如果方法可以转化为没有闭式解的方程 tan x − x = 0,就不能使用根本身。它只能利用二次方程的表面特征——即它的两个系数 2 和 −3。遗憾的是,没有一种可行的将 2 和 −3 结合起来的方法能够预测图像

This brute-force method for computing the root sum requires a solution to the quadratic equation. However, a method that can transfer to the equation tan x − x = 0, which has no closed-form solution, cannot use the roots themselves. It must use only surface features of the quadratic—namely, its two coefficients 2 and −3. Unfortunately, no plausible method of combining 2 and −3 predicts that .

图像 多项式类比哪里出错了?

Where did the polynomial analogy go wrong?

问题在于,二次函数 x 2 − 3x + 2 与 tan x − x 不够相似。二次函数只有正根;然而,奇函数 tan x − x 有对称的正根和负根,并且在 x = 0 处有一个根。事实上,tan x 的泰勒级数为 x + x 3 /3 + 2x 5 /15 + · · ·问题 6.28);因此,

The problem is that the quadratic x2 − 3x + 2 is not sufficiently similar to tan x − x. The quadratic has only positive roots; however, tan x − x, an odd function, has symmetric positive and negative roots and has a root at x = 0. Indeed, the Taylor series for tan x is x + x3/3 + 2x5/15 + · · · (Problem 6.28); therefore,

图片

x 3的公因数意味着 tan x − x 在 x = 0 处有一个三重根。类似的多项式——这里在 x = 0 处有一个三重根,一个正根和一个对称负根——是 (x+2)x 3 (x−2),或者展开后为 x 5 − 4x 3。和式图片(使用正根)只包含一项。并且正好是 1/4。这个值可能是多项式最后两个系数的(负)比值。

The common factor of x3 means that tan x − x has a triple root at x = 0. An analogous polynomial—here, one with a triple root at x = 0, a positive root, and a symmetric negative root—is (x+2)x3(x−2) or, after expansion, x5 − 4x3. The sum (using the positive root) contains only one term and is simply 1/4. This value could plausibly arise as the (negative) ratio of the last two coefficients of the polynomial.

要判断该模式是否是巧合,可以尝试一个更复杂的多项式:其根为 −2、−1、0(三重)、1 和 2。其中一个这样的多项式是

To decide whether that pattern is a coincidence, try a richer polynomial: one with roots at −2, −1, 0 (threefold), 1, and 2. One such polynomial is

图片

多项式根和仅使用两个正根 1 和 2,即 1/1 2 + 1/2 2,等于 5/4——最后两个系数的(负)比。作为对此模式的最后测试,在根中包含 −3 和 3。得到的多项式为

The polynomial-root sum uses only the two positive roots 1 and 2 and is 1/12 + 1/22, which is 5/4—the (negative) ratio of the last two coefficients. As a final test of this pattern, include −3 and 3 among the roots. The resulting polynomial is

图片

多项式根和使用三个正根 1、2 和 3,即 1/1 2 + 1/2 2 + 1/3 2,即 49/36 — 同样是展开多项式中最后两个系数的(负)比率。

The polynomial-root sum uses the three positive roots 1, 2, and 3 and is 1/12 + 1/22 + 1/32, which is 49/36—again the (negative) ratio of the last two coefficients in the expanded polynomial.

图像 这个模式的起源是什么?如何将其扩展到tan x − x

What is the origin of the pattern, and how can it be extended to tan x − x?

为了解释这个模式,将多项式整理如下:

To explain the pattern, tidy the polynomial as follows:

图片

在这个排列中,和 49/36 是第一个值得关注的系数的负数。让我们推广一下。在 x = 0 处取 k 个根,在± x 1 , ± x 2 , . . ., ± x n处取单个根,得到多项式

In this arrangement, the sum 49/36 appears as the negative of the first interesting coefficient. Let’s generalize. Placing k roots at x = 0 and single roots at ±x1, ±x2, . . ., ±xn gives the polynomial

图片

其中 A 为常数。当展开括号内因子的乘积时,展开式中x 2项的系数会从因子中的每个 x 2 /x 2 k项中获得一个贡献。因此,展开式开始于

where A is a constant. When expanding the product of the factors in parentheses, the coefficient of the x2 term in the expansion receives one contribution from each x2/x2k term in a factor. Thus, the expansion begins

图片

括号中的 x 2的系数图片是正切根和的多项式类比。

The coefficient of x2 in parentheses is which is the polynomial analog of the tangent-root sum.

让我们将这种方法应用于 tan x − x。虽然它不是多项式,但它的泰勒级数就像一个无限次多项式。泰勒级数是

Let’s apply this method to tan x − x. Although it is not a polynomial, its Taylor series is like an infinite-degree polynomial. The Taylor series is

图片

x 2系数的负数应为 − 图片。因此,对于正切和问题,图片应为 −2/5 。不幸的是,正量的和不能为负!

The negative of the x2 coefficient should be −. For the tangent-sum problem, should therefore be −2/5. Unfortunately, the sum of positive quantities cannot be negative!

图像 这个类比哪里出了问题?

What went wrong with the analogy?

一个问题是,tan x − x 可能有虚数根或复数根,这些根的平方对 S 的贡献为负。幸运的是,它的所有根都是实数(问题 6.29)。一个更难解决的问题是,tan x − x 在有限的 x 值处趋于无穷大,并且无限次地这样做,而没有一个多项式会这样做,哪怕只有一次。

One problem is that tan x − x might have imaginary or complex roots whose squares contribute negative amounts to S. Fortunately, all its roots are real (Problem 6.29). A harder-to-solve problem is that tan x − x goes to infinity at finite values of x, and does so infinitely often, whereas no polynomial does so even once.

解决方法是构造一个没有无穷大但根与 tan x-x 相同的函数。tan x - x 的无穷大出现在 tan x 爆炸的地方,也就是 cos x = 0 的地方。为了在不产生或消除任何根的情况下消除无穷大,将 tan x - x 乘以 cos x。因此,要展开的类多项式函数是 sin x - x cos x。

The solution is to construct a function having no infinities but having the same roots as tan x−x. The infinities of tan x − x occur where tan x blows up, which is where cos x = 0. To remove the infinities without creating or destroying any roots, multiply tan x − x by cos x. The polynomial-like function to expand is therefore sin x − x cos x.

图片

其泰勒展开式为

Its Taylor expansion is

图片

这两个系列的区别在于

The difference of the two series is

图片

x 3 /3 因子表示 x = 0 处的三重根。最后,作为 x 2系数的负数,就是我们的正切根和 S = 1/10。

The x3/3 factor indicates the triple root at x = 0. And there at last, as the negative of the x2 coefficient, sits our tangent-root sum S = 1/10.

问题 6.28 正切的泰勒级数

Problem 6.28 Taylor series for the tangent

使用 sin x 和 cos x 的泰勒级数来证明

Use the Taylor series for sin x and cos x to show that

图片

提示:使用时取出大的部分。

Hint: Use taking out the big part.

问题 6.29 只有实根

Problem 6.29 Only real roots

证明 tan x − x 的所有根都是实数。

Show that all roots of tan x − x are real.

问题 6.30 精确巴塞尔和

Problem 6.30 Exact Basel sum

使用多项式类比来评估巴塞尔和

Use the polynomial analogy to evaluate the Basel sum

图片

将您的结果与问题 6.22 的解决方案进行比较。

Compare your result with your solution to Problem 6.22.

问题 6.31 误导性的替代扩展

Problem 6.31 Misleading alternative expansions

求tan x = x的平方并取倒数,可得cot 2 x = x −2;等价地,cot 2 x − x −2 = 0。因此,如果x是tan x − x的根,那么它也是cot 2 x − x −2的根。cot 2 x − x −2的泰勒展开式为

Squaring and taking the reciprocal of tan x = x gives cot2 x = x−2; equivalently, cot2 x − x−2 = 0. Therefore, if x is a root of tan x − x, it is a root of cot2 x − x−2. The Taylor expansion of cot2 x − x−2 is

图片

因为 x 2的系数是 −1/10,所以 cot x = x −2的正切根和 S(因此 tan x = x)应该是 1/10。正如我们通过实验和解析计算得出的 tan x = x 的结论一样,这个结论是正确的。但是,这个推理到底错在哪里呢?

Because the coefficient of x2 is −1/10, the tangent-root sum S—for cot x = x−2 and therefore tan x = x—should be 1/10. As we found experimentally and analytically for tan x = x, the conclusion is correct. However, what is wrong with the reasoning?

问题 6.32 倒数的四次方

Problem 6.32 Fourth powers of the reciprocals

sin x − x cos x 的泰勒级数继续

The Taylor series for sin x − x cos x continues

图片

因此,求出图片tan x = x的正根。用数值方法检验你的结果是否合理。

Therefore find for the positive roots of tan x = x. Check numerically that your result is plausible.

问题 6.33 根的其他源方程

Problem 6.33 Other source equations for the roots

找出图片x n是 cos x 的正根。

Find where the xn are the positive roots of cos x.

6.5 一路顺风

6.5 Bon voyage

希望您喜欢将街头格斗技巧融入您的问题解决工具箱。希望您能找到各种机会运用量纲分析、简单案例、归并法、图像推理、去重法和类比法。随着您运用这些工具,您会不断磨练它们,甚至创造出新的工具。

I hope that you have enjoyed incorporating street-fighting methods into your problem-solving toolbox. May you find diverse opportunities to use dimensional analysis, easy cases, lumping, pictorial reasoning, taking out the big part, and analogy. As you apply the tools, you will sharpen them—and even build new tools.

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指数

Index

斜体页码表示该页上的问题。

An italic page number refers to a problem on that page.

ν

ν

参见运动粘度

see kinematic viscosity

1个或几个

1 or few

很少见到

see few

≈(约等于)6

≈ (approximately equal) 6

π,计算

π, computing

反正切级数64

arctangent series 64

布伦特-萨拉明算法65

Brent–Salamin algorithm 65

∝(与……成比例)6

∝ (proportional to) 6

∼ (旋转6,44

∼ (twiddle) 6, 44

ω

ω

参见角频率

see angular frequency

类比推理99–121

analogy, reasoning by 99–121

用平面103–107划分空间

dividing space with planes 103–107

产生猜想

generating conjectures

参见猜想:生成

see conjectures: generating

操作员107–113

operators 107–113

左移 (L) 108–109

left shift (L) 108–109

求和(Σ)109

summation (Σ) 109

保留关键特征100、118、120

preserving crucial features 100, 118, 120

金字塔第19卷

pyramid volume 19

空间角度99–103

spatial angles 99–103

正切根和118–121

tangent-root sum 118–121

测试猜想

testing conjectures

参见猜想:测试

see conjectures: testing

多项式118–121

to polynomials 118–121

变换因变量101

transforming dependent variable 101

角度,空间99–103

angles, spatial 99–103

角频率44

angular frequency 44

亚里士多德十四

Aristotle xiv

算术几何平均值65

arithmetic–geometric mean 65

算术平均数-几何平均数不等式60–66

arithmetic-mean–geometric-mean inequality 60–66

应用63–66

applications 63–66

计算π 64–66

computing π 64–66

最大值63–64

maxima 63–64

平等条件62

equality condition 62

数值算例60

numerical examples 60

图样61–63

pictorial proof 61–63

象征性证明61

symbolic proof 61

算术平均值

arithmetic mean

另请参阅几何平均数

see also geometric mean

62张图片

picture for 62

tan x 114的渐近线

asymptotes of tan x 114

大气压34

atmospheric pressure 34

粗略估计

back-of-the-envelope estimates

纠正78

correcting 78

77心算乘法

mental multiplication in 77

78所需的最低准确度

minimal accuracy required for 78

78中的 10 的幂

powers of 10 in 78

平衡41

balancing 41

巴塞尔和 (∑n −2 ) 76 , 113 , 116 , 121

Basel sum (∑n−2) 76, 113, 116, 121

β函数98

beta function 98

很大一部分,纠正

big part, correcting the

另见取出大部分

see also taking out the big part

加法校正比乘法校正更混乱80

additive messier than multiplicative corrections 80

使用乘法修正

using multiplicative corrections

看到分数变化

see fractional changes

使用一个或几个78

using one or few 78

很大一部分,取出

big part, taking out

看到取出大部分

see taking out the big part

二项式系数96 , 107

binomial coefficients 96, 107

二项分布98

binomial distribution 98

二项式定理90 , 97

binomial theorem 90, 97

平分三角形70–73

bisecting a triangle 70–73

位,CD容量为78

bits, CD capacity in 78

黑体辐射87

blackbody radiation 87

边界层27

boundary layers 27

大脑进化57

brain evolution 57

白金汉,埃德加26

Buckingham, Edgar 26

微积分的基本思想31

calculus, fundamental idea of 31

光盘

CD-ROM

另请参阅 CD

see also CD

与 CD 77格式相同

same format as CD 77

CD/CD-ROM,存储容量77–79

CD/CD-ROM, storage capacity 77–79

特征震级(典型震级)44

characteristic magnitudes (typical magnitudes) 44

特征时间44

characteristic times 44

检查单位78

checking units 78

圆圈

circle

周长76的面积

area from circumference 76

多边形有多条边72

as polygon with many sides 72

比较,不同的废话

comparisons, nonsense with different

尺寸2

dimensions 2

锥体自由落体距离35

cone free-fall distance 35

锥体模板21

cone templates 21

圆锥摆48

conical pendulum 48

猜想

conjectures

丢弃巧合105,119

discarding coincidences 105, 119

解释119

explaining 119

生成100 , 103 , 104 , 105

generating 100, 103, 104, 105

概率为105

probabilities of 105

测试100 , 101 , 104 , 106 , 111 , 119

testing 100, 101, 104, 106, 111, 119

获取更多数据100 , 105 , 106

getting more data 100, 105, 106

比例常数

constants of proportionality

斯特藩-玻尔兹曼常数11

Stefan–Boltzmann constant 11

约束传播5

constraint propagation 5

矛盾20

contradictions 20

收敛,加速65 , 68

convergence, accelerating 65, 68

凸度104

convexity 104

版权提高书价82

copyright raising book prices 82

科菲尔德,大卫105

Corfield, David 105

余弦

cosine

高次积分94–97

integral of high power 94–97

小角度近似推导86

small-angle approximation derived 86

二手95

used 95

立方体,平分73

cube, bisecting 73

d(差分符号)10,43

d (differential symbol) 10, 43

退化103

degeneracies 103

导数作为比率38

derivative as a ratio 38

衍生品

derivatives

用非零 Δx 40近似

approximating with nonzero Δx 40

割线近似38

secant approximation 38

39中的错误

errors in 39

改进的起点39

improved starting point 39

大错误38

large error 38

垂直平移39

vertical translation 39

第二

second

尺寸为38

dimensions of 38

割线近似为38

secant approximation to 38

显著变化近似值40–41

significant-change approximation 40–41

加速度43

acceleration 43

纳维-斯托克斯导数45

Navier–Stokes derivatives 45

尺度和平移不变性40

scale and translation invariance 40

平移不变性40

translation invariance 40

荒岛法32

desert-island method 32

微分方程

differential equations

检查尺寸42

checking dimensions 42

线性化47 , 51–54

linearizing 47, 51–54

轨道运动12

orbital motion 12

摆锤46

pendulum 46

简化为代数方程43–46

simplifying into algebraic equations 43–46

弹簧质量系统42–45

spring–mass system 42–45

精确解45

exact solution 45

钟摆方程47

pendulum equation 47

维度分析

dimensional analysis

参见维度、方法;无量纲群

see dimensions, method of; dimensionless groups

无量纲常数

dimensionless constants

高斯积分10

Gaussian integral 10

简谐运动48

simple harmonic motion 48

斯特藩-玻尔兹曼定律11

Stefan–Boltzmann law 11

无量纲群24

dimensionless groups 24

拖拽25

drag 25

自由落体速度24

free-fall speed 24

钟摆周期48

pendulum period 48

弹簧质量系统48

spring–mass system 48

无量纲量

dimensionless quantities

94号井深度

depth of well 94

分数变化乘以指数89

fractional change times exponent 89

熵值较低94

have lower entropy 94

熵值较低81

having lower entropy 81

方面

dimensions

L 代表长度5

L for length 5

保留5

retaining 5

T 代表时间5

T for time 5

与单位2相比

versus units 2

尺寸, 1-12方法

dimensions, method of 1–12

另见无量纲群

see also dimensionless groups

优势6

advantages 6

检查微分方程42

checking differential equations 42

选择未指定的维度7、8–9

choosing unspecified dimensions 7, 8–9

与简单案例相比15

compared with easy cases 15

约束传播5

constraint propagation 5

拖拽23–26

drag 23–26

猜测积分7–11

guessing integrals 7–11

开普勒第三定律12

Kepler’s third law 12

钟摆48–49

pendulum 48–49

相关利率问题12

related-rates problems 12

求解微分方程的稳健替代方法5

robust alternative to solving differential equations 5

斯特藩-玻尔兹曼定律11

Stefan–Boltzmann law 11

尺寸

dimensions of

角度47

angles 47

d(微分)10

d (differential) 10

dχ10

10

指数8

exponents 8

积分9

integrals 9

积分符号 ∫ 9

integration sign ∫ 9

运动粘度ν22

kinematic viscosity ν 22

钟摆方程47

pendulum equation 47

二阶导数38 , 43

second derivative 38, 43

弹簧常数43

spring constant 43

求和符号 Σ 9

summation sign Σ 9

拖拽21–29

drag 21–29

井深估计,对93的影响

depth-of-well estimate, effect on 93

高雷诺数28

high Reynolds number 28

低雷诺数30

low Reynolds number 30

影响23的数量

quantities affecting 23

阻力

drag force

参见阻力

see drag

e

e

分数变化90

in fractional changes 90

地球

earth

表面积79

surface area 79

表面温度87

surface temperature 87

简单案例13–30

easy cases 13–30

奇数加法58

adding odd numbers 58

β函数积分98

beta-function integral 98

平分三角形70

bisecting a triangle 70

键角100

bond angles 100

检查公式13–17

checking formulas 13–17

与尺寸15相比

compared with dimensions 15

椭圆区域16–17

ellipse area 16–17

椭圆周长65

ellipse perimeter 65

更少的行数104

fewer lines 104

飞机数量减少103

fewer planes 103

猜测积分13–16

guessing integrals 13–16

高维103

high dimensionality 103

高雷诺数27

high Reynolds number 27

大指数89

large exponents 89

低雷诺数30

low Reynolds number 30

无限声速92 , 94

of infinite sound speed 92, 94

pendulum

大振幅49–51

large amplitude 49–51

小振幅47–48

small amplitude 47–48

多项式118

polynomials 118

金字塔第19卷

pyramid volume 19

tan χ = χ 114的根

roots of tan χ = χ 114

简单函数108 , 112

simple functions 108, 112

合成公式17

synthesizing formulas 17

截头圆锥21

truncated cone 21

截锥形金字塔18–21

truncated pyramid 18–21

椭圆

ellipse

17

area 17

周长65

perimeter 65

椭圆轨道

elliptical orbit

偏心率87

eccentricity 87

太阳的位置87

position of sun 87

节能50

energy conservation 50

驾驶过程中的能源消耗82–84

energy consumption in driving 82–84

通勤时间延长的影响

effect of longer commuting time 83

表达式的熵

entropy of an expression

参见低熵表达式

see low-entropy expressions

混合熵81

entropy of mixing 81

平等, 6

equality, kinds of 6

估算衍生品

estimating derivatives

参见导数,正割近似;导数,显著变化近似

see derivatives, secant approximation; derivatives, significant-change approximation

欧拉113

Euler 113

另请参阅巴塞尔协议

see also Basel sum

β函数98

beta function 98

欧拉-麦克劳林求和112

Euler–MacLaurin summation 112

进化中的大脑 57

Evolving Brains 57

精确解

exact solution

引发代数错误4

invites algebra mistakes 4

示例

examples

奇数加法58–60

adding odd numbers 58–60

算术平均数-几何平均数不等式60–66

arithmetic-mean–geometric-mean inequality 60–66

婴儿数量:32-33

babies, number of 32–33

平分三角形70–73

bisecting a triangle 70–73

甲烷的键角99–103

bond angle in methane 99–103

井深91–94

depth of a well 91–94

cos x 的导数,估计40–41

derivative of cos x, estimating 40–41

用平面103–107划分空间

dividing space with planes 103–107

拖拽下落的纸锥21–29

drag on falling paper cones 21–29

椭圆区域16–17

ellipse area 16–17

55 英里/小时限速下的节能效果82–84

energy savings from 55 mph speed limit 82–84

阶乘函数36–37

factorial function 36–37

自由落体3–6

free fall 3–6

使用维度7–11 的高斯积分

Gaussian integral using dimensions 7–11

使用简单情况13–16 的高斯积分

Gaussian integral using easy cases 13–16

对数级数66–70

logarithm series 66–70

最大化花园面积63–64

maximizing garden area 63–64

将 3.15 乘以 7.21

multiplying 3.15 by 7.21

使用分数变化79–80

using fractional changes 79–80

使用一个或几个79

using one or few 79

运算符

operators

左移 (L) 108–109

left shift (L) 108–109

求和(Σ)109–113

summation (Σ) 109–113

钟摆周期46–54

pendulum period 46–54

跨国公司的力量1–3

power of multinationals 1–3

快速计算 1/13 84–85

rapidly computing 1/13 84–85

季节性温度波动86–88

seasonal temperature fluctuations 86–88

弹簧质量微分方程42–45

spring–mass differential equation 42–45

十的平方根85–86

square root of ten 85–86

CD-ROM 或 CD 的存储容量77–79

storage capacity of a CD-ROM or CD 77–79

求和 ln n! 73–75

summing ln n! 73–75

正切根和113–121

tangent-root sum 113–121

三角积分94–97

trigonometric integral 94–97

截头金字塔的体积17–21

volume of truncated pyramid 17–21

指数

exponential

衰减,33的积分

decaying, integral of 33

超越任何多项式36

outruns any polynomial 36

指数,维度为8

exponents, dimensions of 8

极端情况

extreme cases

参见简单案例

see easy cases

阶乘

factorial

积分表示36

integral representation 36

斯特林公式

Stirling’s formula

欧拉-麦克劳林求和112

Euler–MacLaurin summation 112

36–37

lumping 36–37

图片74

pictures 74

求和表示73

summation representation 73

73–75的求和对数

summing logarithm of 73–75

很少

few

几何平均数

as geometric mean 78

发明数字78

as invented number 78

心算乘法78

for mental multiplication 78

分数变化

fractional changes

立方根86

cube roots 86

立方83 , 84

cubing 83, 84

不要繁殖

do not multiply 83

地球与太阳的距离87

earth–sun distance 87

估算风力发电量84

estimating wind power 84

−2 的指数86

exponent of −2 86

1/4 的指数87

exponent of 1/4 87

一般指数84–90

general exponents 84–90

提高准确度85 , 86

increasing accuracy 85, 86

引入79–80

introduced 79–80

大指数89–90 , 95

large exponents 89–90, 95

线性近似82

linear approximation 82

3.15 乘以 7.21 79

multiplying 3.15 by 7.21 79

负指数和分数指数86–88

negative and fractional exponents 86–88

除了增加82之外没有其他可行的替代方案

no plausible alternative to adding 82

图片80

picture 80

小改动添加82

small changes add 82

平方根85–86

square roots 85–86

平方82–84

squaring 82–84

正切根和117

tangent-root sum 117

自由落体

free fall

使用维度3-6进行分析

analysis using dimensions 3–6

91-94井深度

depth of well 91–94

微分方程4

differential equation 4

撞击速度(精确)4

impact speed (exact) 4

初速度30

with initial velocity 30

捏造33

fudging 33

燃油效率85

fuel efficiency 85

高斯积分

Gaussian integral

闭式,猜测14 , 16

closed form, guessing 14, 16

将极限扩展到 96

extending limits to 96

尾部区域55

tail area 55

梯形近似14

trapezoidal approximation 14

使用尺寸7–11

using dimensions 7–11

使用简单的案例13–16

using easy cases 13–16

使用集中34 , 35

using lumping 34, 35

GDP,作为货币流量1

GDP, as monetary flow 1

几何平均值

geometric mean

另请参阅算术平均值;算术平均数-几何平均数定理

see also arithmetic mean; arithmetic-mean–geometric-mean theorem

定义60

definition 60

61张照片

picture for 61

三个数字63

three numbers 63

格式塔理解59

gestalt understanding 59

全球化1

globalization 1

图形论证参见图解证明

graphical arguments see pictorial proofs

高熵表达式

high-entropy expressions

另见低熵表达式

see also low-entropy expressions

从二次公式92

from quadratic formula 92

如何解决它 xiii

How to Solve It xiii

惠更斯48

Huygens 48

感应证明58

induction proof 58

信息论81

information theory 81

一体化

integration

近似为乘法,参见集中

approximating as multiplication see lumping

微分逆109

inverse of differentiation 109

数字14

numerical 14

操作员109

operator 109

太阳辐射强度86

intensity of solar radiation 86

等周定理73

isoperimetric theorem 73

杰恩斯,埃德温·汤普森105

Jaynes, Edwin Thompson 105

杰弗里斯,哈罗德26

Jeffreys, Harold 26

开普勒第三定律25

Kepler’s third law 25

运动粘度(v 21,27

kinematic viscosity (v) 21, 27

朗道研究所,令人畏惧的三角积分从94年开始

Landau Institute, daunting trigonometric integral from 94

L(长度尺寸)5

L (dimension of length) 5

伦纳德-琼斯潜力41

Lennard–Jones potential 41

预期寿命32

life expectancy 32

一点点(d 的意思10,43

little bit (meaning of d) 10, 43

对数

logarithms

分析分数变化90

analyzing fractional changes 90

积分定义67

integral definition 67

有理函数近似69

rational-function approximation 69

低熵表达式

low-entropy expressions

科学进步的基础

basis of scientific progress 81

无量纲量通常为81

dimensionless quantities are often 81

分数变化往往

fractional changes are often 81

逐次逼近法93

from successive approximation 93

高熵中间步骤81

high-entropy intermediate steps 81

引入80–82

introduced 80–82

降低混合熵81

reducing mixing entropy 81

tan x = x 114的根

roots of tan x = x 114

集结31–55

lumping 31–55

1/e启发式34

1/e heuristic 34

大气压34

atmospheric pressure 34

外接矩形67

circumscribed rectangle 67

微分方程51–54

differential equations 51–54

估算导数37–41

estimating derivatives 37–41

内接矩形67

inscribed rectangle 67

积分33–37

integrals 33–37

摆,中等振幅51

pendulum, moderate amplitudes 51

人口估计数32–33

population estimates 32–33

太多

too much 52

火星气候探测器坠毁,3架

Mars Climate Orbiter, crash of 3

数学与合理推理 xiii

Mathematics and Plausible Reasoning xiii

数学,抽象能力7

mathematics, power of abstraction 7

最大值和最小值41 , 70

maxima and minima 41, 70

算术平均数-几何平均数不等式63–64

arithmetic-mean–geometric-mean inequality 63–64

盒装容量64

box volume 64

三角学64

trigonometry 64

心理分裂33

mental division 33

心算乘法

mental multiplication

使用一个或几个

using one or few

很少见到

see few

方法与技巧69

method versus trick 69

混合熵81

mixing entropy 81

纳维-斯托克斯方程

Navier–Stokes equations

难以解决22

difficult to solve 22

惯性项45

inertial term 45

21日声明

statement of 21

粘性术语46

viscous term 46

牛顿-拉夫森方法76 , 117 , 118

Newton–Raphson method 76, 117, 118

数值积分14

numerical integration 14

奇数,58–60的和

odd numbers, sum of 58–60

一个或几个

one or few

如果不够准确

if not accurate enough 79

运算符

operators

衍生物(D)107

derivative (D) 107

108的指数

exponential of 108

有限差分(Δ)110

finite difference (Δ) 110

整合109

integration 109

左移 (L) 108–109

left shift (L) 108–109

右移109

right shift 109

求和(Σ)109–113

summation (Σ) 109–113

抛物线,面积(无需微积分)76

parabola, area without calculus 76

帕斯卡三角形107

Pascal’s triangle 107

模式,寻找90

patterns, looking for 90

pendulum

微分方程46

differential equation 46

在较弱的重力

in weaker gravity 52

46-54

period of 46–54

感知能力58

perceptual abilities 58

图样57–76

pictorial proofs 57–76

奇数加法58–60

adding odd numbers 58–60

圆的面积76

area of circle 76

算术平均几何平均不等式60–63 , 76

arithmetic-mean–geometric-mean inequality 60–63, 76

平分三角形70–73

bisecting a triangle 70–73

与归纳证明58相比

compared to induction proof 58

用平面划分空间107

dividing space with planes 107

阶乘73–75

factorial 73–75

对数级数66–70

logarithm series 66–70

牛顿-拉夫森方法76

Newton–Raphson method 76

tan x = x 114的根

roots of tan x = x 114

球体体积76

volume of sphere 76

图形推理

pictorial reasoning

94号井深度

depth of well 94

可行的替代方案

plausible alternatives

参见低熵表达式

see low-entropy expressions

波利亚,乔治105

Polya, George 105

人口估计为32

population, estimating 32

跨国公司的力量1–3

power of multinationals 1–3

十的幂78

powers of ten 78

比例推理18

proportional reasoning 18

金字塔,截顶17

pyramid, truncated 17

二次公式91

quadratic formula 91

高熵92

high entropy 92

与逐次逼近法93

versus successive approximation 93

二次项

quadratic terms

忽略808284

ignoring 80, 82, 84

包括85

including 85

范围公式30

range formula 30

快速心理分裂84–85

rapid mental division 84–85

有理函数69 , 101

rational functions 69, 101

关于

Re

参见雷诺数

see Reynolds number

相关利率问题12

related-rates problems 12

重写比例技巧68 , 70 , 86

rewriting-as-a-ratio trick 68, 70, 86

雷诺数(Re)27

Reynolds number (Re) 27

27

high 27

30

low 30

严谨十三

rigor xiii

尸僵 xiii

rigor mortis xiii

四舍五入

rounding

精确到最接近的整数79

to nearest integer 79

使用一个或几个78

using one or few 78

尺度不变性40

scale invariance 40

季节性温度变化86–88

seasonal temperature changes 86–88

季节性气温波动

seasonal temperature fluctuations

替代解释88

alternative explanation 88

割线近似

secant approximation

参见导数、正割近似

see derivatives, secant approximation

割线,斜率为38

secant line, slope of 38

二阶导数

second derivatives

参见衍生品,第二

see derivatives, second

香农-奈奎斯特采样定理78

Shannon–Nyquist sampling theorem 78

显著变化近似值

significant-change approximation

参见衍生品,重大变化

see derivatives, significant-change

近似

approximation

相似三角形61 , 70

similar triangles 61, 70

简化问题

simplifying problems

参见取出大部分;集中;简单案例;类比

see taking out the big part; lumping; easy cases; analogy

正弦,小角度近似

sine, small-angle approximation

衍生47

derived 47

使用86

used 86

小角度近似

small-angle approximation

余弦95

cosine 95

正弦47 , 66

sine 47, 66

太阳辐射强度86

solar-radiation intensity 86

空间,用平面103–107划分

space, dividing with planes 103–107

光谱学35

spectroscopy 35

球体,表面积76计算体积

sphere, volume from surface area 76

弹簧质量系统42–45

spring–mass system 42–45

弹簧常数

spring constant

尺寸为43

dimensions of 43

胡克定律,42

Hooke’s law, in 42

统计力学81

statistical mechanics 81

斯特藩-玻尔兹曼常数11 , 87

Stefan–Boltzmann constant 11, 87

斯特藩-玻尔兹曼定律

Stefan–Boltzmann law

推导11

derivation 11

要求温度为88开尔文

requires temperature in Kelvin 88

计算表面温度87

to compute surface temperature 87

刚性

stiffness

参见弹簧常数

see spring constant

斯特林公式

Stirling’s formula

参见阶乘:斯特林公式

see factorial: Stirling’s formula

逐次逼近法

successive approximation

另见取出大部分

see also taking out the big part

92–94井深度

depth of well 92–94

低熵表达式93

low-entropy expressions 93

物理洞察93

physical insights 93

稳健性93

robustness 93

与二次公式93

versus quadratic formula 93

总结

summation

近似积分113 , 114

approximately integration 113, 114

欧拉-麦克劳林112 , 113

Euler–MacLaurin 112, 113

不确定110

indefinite 110

积分近似74

integral approximation 74

操作员109–113

operator 109–113

使用微分表示112

represented using differentiation 112

切线根113–121

tangent roots 113–121

三角校正74 , 113 , 115

triangle correction 74, 113, 115

符号推理

symbolic reasoning

大脑进化57

brain evolution 57

看起来像魔术61

seeming like magic 61

对称性72

symmetry 72

取出大部分77–98

taking out the big part 77–98

92–94井深度

depth of well 92–94

多项式外推106 , 107

polynomial extrapolation 106, 107

正切根和114 , 117–118

tangent-root sum 114, 117–118

三角积分94–97

trigonometric integral 94–97

泰勒级数

Taylor series

阶乘被积函数37

factorial integrand 37

一般66

general 66

对数66 , 69

logarithm 66, 69

三次项68

cubic term 68

钟摆周期53

pendulum period 53

切线118 , 120

tangent 118, 120

L(长度尺寸)5

L (dimension of length) 5

四面体,正99

tetrahedron, regular 99

解决问题的艺术与技巧 xiii

The Art and Craft of Problem Solving xiii

热膨胀82

thermal expansion 82

汤普森,西尔瓦努斯10

Thompson, Silvanus 10

思想实验18 , 50

thought experiments 18, 50

工具

tools

参见维度、方法;简单案例;集中;图形证明;取出大部分;类比、推理

see dimensions, method of; easy cases; lumping; pictorial proofs; taking out the big part; analogy, reasoning by

转型

transformations

对数36

logarithmic 36

取余弦101

taking cosine 101

梯形近似14

trapezoidal approximation 14

技巧

tricks

乘以一85

multiplication by one 85

重写为比率68 , 70 , 86

rewriting as a ratio 68, 70, 86

变量变换36 , 101

variable transformation 36, 101

技巧与方法69

trick versus method 69

辅导教学xiv

tutorial teaching xiv

低估还是高估?

under- or overestimate?

92、93井的近似深度

approximating depth of well 92, 93

计算平方根86

computing square roots 86

集中分析54

lumping analysis 54

求和近似75

summation approximation 75

正切根和115

tangent-root sum 115

使用一个或几个79

using one or few 79

单位

units

取消78

cancellation 78

火星气候探测器坠毁,3架

Mars Climate Orbiter, crash of 3

与数量4分离

separating from quantities 4

与维度2相比

versus dimensions 2

韦特海默,马克斯59

Wertheimer, Max 59

本书完全使用免费软件和字体创建。文本采用赫尔曼·察普夫(Hermann Zapf)设计的Palatino语言编写,并以TeX ​​Gyre Pagella格式提供。数学部分采用欧拉语言编写,欧拉语言也由赫尔曼·察普夫设计。

This book was created entirely with free software and fonts. The text is set in Palatino, designed by Hermann Zapf and available as TeX Gyre Pagella. The mathematics is set in Euler, also designed by Hermann Zapf.

Maxima 5.17.1 和 mpmath Python 库辅助进行了几项计算。

Maxima 5.17.1 and the mpmath Python library aided several calculations.

源文件使用多个版本的 GNU Emacs 创建,并使用 Mercurial 版本控制系统进行管理。图形源文件使用 MetaPost 1.208 和 Asymptote 1.88 编译。TEX 源文件使用 ConTeXt 2009.10.27 和 PDFTeX 1.40.10 编译为 PDF。编译过程使用 GNU Make 3.81 进行管理,在一台 2006 年的老式笔记本电脑上耗时 10 分钟。所有软件均在 Debian GNU/Linux 上运行。

The source files were created using many versions of GNU Emacs and managed using the Mercurial revision-control system. The figure source files were compiled with MetaPost 1.208 and Asymptote 1.88. The TEX source was compiled to PDF using ConTeXt 2009.10.27 and PDFTeX 1.40.10. The compilations were managed with GNU Make 3.81 and took 10 min on a 2006-vintage laptop. All software was running on Debian GNU/Linux.

我热烈感谢软件共享的众多贡献者。

I warmly thank the many contributors to the software commons.